Multivariable calculus integration

Multivariable calculus integration DEFAULT

Double Integrals Over Rectangles

For a rectangular region [latex]S[/latex] defined by [latex]x[/latex] in [latex][a,b][/latex] and [latex]y[/latex] in [latex][c,d][/latex], the double integral of a function [latex]f(x,y)[/latex] in this region is given as [latex]\int_c^d(\int_a^b f(x,y) dx) dy[/latex].

Learning Objectives

Use double integrals to find the volume of rectangular regions in the xy-plane

Key Takeaways

Key Points

  • The multiple integral is a type of definite integral extended to functions of more than one real variable —for example, [latex]f(x, y)[/latex] or [latex]f(x, y, z)[/latex]. Integrals of a function of two variables over a region in [latex]R^2[/latex] are called double integrals.
  • The double integral of a positive function of two variables represents the volume of the region between the surface defined by the function (on the three dimensional Cartesian plane where [latex]z = f(x, y))[/latex] and the plane which contains its domain.
  • If there are more variables than 3, a multiple integral will yield hypervolumes of multi-dimensional functions.

Key Terms

  • Fubini’s theorem: a result which gives conditions under which it is possible to compute a double integral using iterated integrals
  • hypervolume: a volume in more than three dimensions

The multiple integral is a type of definite integral extended to functions of more than one real variable—for example, [latex]f(x, y)[/latex] or [latex]f(x, y, z)[/latex]. Integrals of a function of two variables over a region in [latex]R^2[/latex] are called double integrals. Just as the definite integral of a positive function of one variable represents the area of the region between the graph of the function and the [latex]x[/latex]-axis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function (on the three dimensional Cartesian plane where [latex]z = f(x, y))[/latex] and the plane which contains its domain. The same volume can be obtained via the triple integral—the integral of a function in three variables—of the constant function [latex]f(x, y, z) = 1[/latex] over the above-mentioned region between the surface and the plane. If there are more variables, a multiple integral will yield hypervolumes of multi-dimensional functions.

image

Volume to be Integrated: Double integral as volume under a surface [latex]z = x^2 − y^2[/latex]. The rectangular region at the bottom of the body is the domain of integration, while the surface is the graph of the two-variable function to be integrated.

Double Integrals Over Rectangles

Double integrals over rectangular regions are straightforward to compute in many cases. For a rectangular region [latex]S[/latex] defined by [latex]x[/latex] in [latex][a,b][/latex] and [latex]y[/latex] in [latex][c,d][/latex], the double integral of a function [latex]f(x,y)[/latex] in this region is given as:

[latex]\begin{align}\int\!\!\!\int_S f(x,y) dxdy &= \int_a^b\left(\int_c^d f(x,y) dy\right) dx \\ &= \int_c^d\left(\int_a^b f(x,y) dx\right) dy\end{align}[/latex].

Here, we exchanged the order of the integration, assuming that [latex]f(x,y)[/latex] satisfies the conditions to apply Fubini’s theorem.

Example

Let us assume that we wish to integrate a multivariable function [latex]f[/latex] over a region [latex]A[/latex]:

[latex]A = \left \{ (x,y) \in \mathbf{R}^2: 11 \le x \le 14 \; \ 7 \le y \le 10 \right \}[/latex]

[latex]f(x,y) = x^2 + 4y[/latex]

Formulating the double integral, we first evaluate the inner integral with respect to [latex]x[/latex]:

[latex]\begin{align} \int_{11}^{14} (x^2 + 4y) \ dx & = \left (\frac{1}{3}x^3 + 4yx \right)\Big |_{x=11}^{x=14} \\ & = \frac{1}{3}(14)^3 + 4y(14) - \frac{1}{3}(11)^3 - 4y(11) \\ &= 471 + 12y \end{align}[/latex]

We then integrate the result with respect to [latex]y[/latex]:

[latex]\begin{align} \int_7^{10} (471 + 12y) \ dy & = (471y + 6y^2)\big |_{y=7}^{y=10} \\ & = 471(10)+ 6(10)^2 - 471(7) - 6(7)^2 \\ &= 1719 \end{align}[/latex]

We could have computed the double integral starting from the integration over [latex]y[/latex]. Confirm yourself that the result is the same.

Iterated Integrals

An iterated integral is the result of applying integrals to a function of more than one variable.

Learning Objectives

Use iterated integrals to integrate a function with more than one variable

Key Takeaways

Key Points

  • The function [latex]f(x,y)[/latex], if [latex]y[/latex] is considered a given parameter, can be integrated with respect to [latex]x[/latex] as follows: [latex]\int f(x,y)dx[/latex].
  • The result is a function of [latex]y[/latex] and therefore its integral can be considered again. If this is done, the result is the iterated integral [latex]\int\left(\int f(x,y)\,dx\right)\,dy[/latex].
  • It is key to note that this is different, in principle, to the multiple integral [latex]\iint f(x,y)\,dx\,dy[/latex].

Key Terms

  • Fubini’s theorem: a result which gives conditions under which it is possible to compute a double integral using iterated integrals

An iterated integral is the result of applying integrals to a function of more than one variable (for example [latex]f(x,y)[/latex] or [latex]f(x,y,z)[/latex]) in such a way that each of the integrals considers some of the variables as given constants. For example, in the function [latex]f(x,y)[/latex], if [latex]y[/latex] is considered a given parameter, it can be integrated with respect to [latex]x[/latex], [latex]\int f(x,y)dx[/latex]. The result is a function of y and therefore its integral can be considered again. If this is done, the result is the iterated integral:

[latex]\displaystyle{\int\left(\int f(x,y)\,dx\right)\,dy}[/latex]

It is key to note that this is different, in principle, from the multiple integral [latex]\iint f(x,y)\,dx\,dy[/latex]. A theorem called Fubini’s theorem, however, states that they may be equal under very mild conditions. The alternative notation for iterated integrals [latex]\int dy \int f(x,y)\,dx[/latex] is also used. Iterated integrals are computed following the operational order indicated by the parentheses (in the notation that uses them), starting from the innermost integral and working out.

image

Use of an iterated integral: An iterated integral can be used to find the volume of the object in the figure.

Example

For the iterated integral [latex]\int\left(\int (x+y) \, dx\right) \, dy[/latex], the integral [latex]\int (x+y) \, dx = \frac{x^2}{2} + yx[/latex] is computed first. The result is then used to compute the integral with respect to [latex]y[/latex]:

[latex]\displaystyle{\int \left(\frac{x^2}{2} + yx \right) \, dy = \frac{yx^2}{2} + \frac{xy^2}{2}}[/latex]

It should be noted, however, that this example omits the constants of integration. After the first integration with respect to [latex]x[/latex], we would rigorously need to introduce a “constant” function of [latex]y[/latex]. That is, If we were to differentiate this function with respect to [latex]x[/latex], any terms containing only [latex]y[/latex] would vanish, leaving the original integral. Similarly for the second integral, we would introduce a “constant” function of [latex]x[/latex], because we have integrated with respect to [latex]y[/latex]. In this way, indefinite integration does not make much sense for functions of several variables. While the antiderivatives of single variable functions differ at most by a constant, the antiderivatives of multivariable functions differ by unknown single-variable terms, which could have a drastic effect on the behavior of the function.

Double Integrals Over General Regions

Double integrals can be evaluated over the integral domain of any general shape.

Learning Objectives

Use double integrals to integrate over general regions

Key Takeaways

Key Points

  • If the domain [latex]D[/latex] is normal with respect to the [latex]x[/latex]– axis, and [latex]f:D \to R[/latex] is a continuous function, then [latex]\alpha(x)[/latex]  and [latex]\beta(x)[/latex] (defined on the interval [latex][a, b][/latex]) are the two functions that determine [latex]D[/latex]: [latex]\iint_D f(x,y)\ dx\, dy = \int_a^b dx \int_{ \alpha (x)}^{ \beta (x)} f(x,y)\, dy[/latex]
  • Applying this general method, the projection of [latex]D[/latex] onto either the [latex]x[/latex]-axis or the [latex]y[/latex]-axis should be bounded by the two values, [latex]a[/latex] and [latex]b[/latex].
  • For a domain [latex]D = \{ (x,y) \in \mathbf{R}^2 \: \ x \ge 0, y \le 1, y \ge x^2 \}[/latex], we can write the integral over [latex]D[/latex] as[latex]\iint_D (x+y) \, dx \, dy = \int_0^1 dx \int_{x^2}^1 (x+y) \, dy[/latex].

Key Terms

  • domain: the set of all possible mathematical entities (points) where a given function is defined

We studied how double integrals can be evaluated over a rectangular region. But there is no reason to limit the domain to a rectangular area. The integral domain can be of any general shape. In this atom, we will study how to formulate such an integral.

This method is applicable to any domain [latex]D[/latex] for which:

  • the projection of [latex]D[/latex] onto either the [latex]x[/latex]-axis or the [latex]y[/latex]-axis is bounded by the two values, [latex]a[/latex] and [latex]b[/latex].
  • any line perpendicular to this axis that passes between these two values intersects the domain in an interval whose endpoints are given by the graphs of two functions, [latex]\alpha[/latex] and [latex]\beta[/latex].

[latex]x[/latex]-axis: If the domain [latex]D[/latex] is normal with respect to the [latex]x[/latex]-axis, and [latex]f:D \to R[/latex] is a continuous function, then [latex]\alpha(x)[/latex]  and [latex]\beta(x)[/latex] (defined on the interval [latex][a, b][/latex]) are the two functions that determine [latex]D[/latex]. It follows, then, that:

[latex]\displaystyle{\iint_D f(x,y)\ dx\, dy = \int_a^b dx \int_{ \alpha (x)}^{ \beta (x)} f(x,y)\, dy}[/latex]

[latex]y[/latex]-axis: If [latex]D[/latex] is normal with respect to the [latex]y[/latex]-axis and [latex]f:D \to R[/latex] is a continuous function, then [latex]\alpha(y)[/latex] and [latex]\beta(y)[/latex] (defined on the interval [latex][a, b][/latex]) are the two functions that determine [latex]D[/latex]. It follows, then, that

[latex]\displaystyle{\iint_D f(x,y)\ dx\, dy = \int_a^b dy \int_{\alpha (y)}^{ \beta (y)} f(x,y)\, dx}[/latex]

Example

Consider the following region:

[latex]D = \{ (x,y) \in \mathbf{R}^2 \: \ x \ge 0, y \le 1, y \ge x^2 \}[/latex]

Calculate [latex]\iint_D (x+y) \, dx \, dy[/latex]. This domain is normal with respect to both the [latex]x[/latex]– and [latex]y[/latex]-axes. To apply the formulae, you must first find the functions that determine [latex]D[/latex] and the intervals over which these are defined. In this case the two functions are [latex]\alpha (x) = x^2[/latex] and [latex]\beta (x) = 1[/latex], while the interval is given by the intersections of the functions with [latex]x=0[/latex], so the interval is [latex][a,b] = [0,1][/latex] (normality has been chosen with respect to the [latex]x[/latex]-axis for a better visual understanding).

image

Double Integral: Double integral over the normal region [latex]D[/latex] shown in the example.

It is now possible to apply the formula:

[latex]\begin{align}\iint_D (x+y) \, dx \, dy &= \int_0^1 dx \int_{x^2}^1 (x+y) \, dy \\ &= \int_0^1 dx \ \left[xy + \frac{y^2}{2} \right]^1_{x^2}\end{align}[/latex]

(At first the second integral is calculated considering [latex]x[/latex] as a constant). The remaining operations consist of applying the basic techniques of integration:

[latex]\begin{align}\int_0^1 \left[xy + \frac{y^2}{2}\right]^1_{x^2} \, dx &= \int_0^1 \left(x + \frac{1}{2} - x^3 - \frac{x^4}{2} \right) dx \\ &= \frac{13}{20}\end{align}[/latex]

Double Integrals in Polar Coordinates

When domain has a cylindrical symmetry and the function has several specific characteristics, apply the transformation to polar coordinates.

Learning Objectives

Solve double integrals in polar coordinates

Key Takeaways

Key Points

  • The fundamental relation to make the transformation is the following: [latex]f(x,y) \rightarrow f(\rho \cos \varphi,\rho \sin \varphi )[/latex].
  • To switch the integral from Cartesian to polar coordinates, the [latex]dx \,\, dy[/latex] differentials in this transformation become [latex]\rho \,\, d\rho \,\,d\varphi[/latex].
  • Once the function is transformed and the domain evaluated, it is possible to define the formula for the change of variables in polar coordinates: [latex]\iint_D f(x,y) \ dx\,\, dy = \iint_T f(\rho \cos \varphi, \rho \sin \varphi) \rho \,\, d \rho\,\, d \varphi[/latex].

Key Terms

  • Cartesian: of or pertaining to co-ordinates based on mutually orthogonal axes
  • Jacobian determinant: the determinant of the Jacobian matrix

In [latex]R^2[/latex], if the domain has a cylindrical symmetry and the function has several particular characteristics, you can apply the transformation to polar coordinates, which means that the generic points [latex]P(x, y)[/latex] in Cartesian coordinates switch to their respective points in polar coordinates. This allows one to change the shape of the domain and simplify the operations.

image

Transformation to Polar Coordinates: This figure illustrates graphically a transformation from cartesian to polar coordinates

Change of variable

The polar coordinates [latex]r[/latex] and [latex]\varphi[/latex] can be converted to the Cartesian coordinates [latex]x[/latex] and [latex]y[/latex] by using the trigonometric functions sine and cosine:

[latex]x = r \cos \varphi \, \\ y = r \sin \varphi \,[/latex]

The Cartesian coordinates [latex]x[/latex] and [latex]y[/latex] can be converted to polar coordinates [latex]r[/latex] and [latex]\varphi[/latex] with [latex]r \geq 0[/latex] and [latex]\varphi[/latex] in the interval [latex](−\pi, \pi][/latex]:

[latex]r = \sqrt{x^2 + y^2} \\ \varphi = \tan ^{-1} (\frac yx)[/latex]

The fundamental relation to make the transformation is as follows:

[latex]f(x,y) \rightarrow f(\rho \cos \phi,\rho \sin \phi )[/latex]

Examples

  • Given the function [latex]f(x,y) = x + y[/latex] and applying the transformation, one obtains [latex]f(\rho, \phi) = \rho \cos \phi + \rho \sin \phi = \rho(\cos \phi + \sin \phi )[/latex].
  • Given the function [latex]f(x,y) = x^2 + y^2[/latex], one can obtain [latex]f(\rho, \phi) = \rho^2 (\cos^2 \phi + \sin^2 \phi) = \rho^2[/latex] using the Pythagorean trigonometric identity, which is very useful to simplify this operation. Particularly in this case, you can see that the representation of the function f became simpler in polar coordinates. This is the case because the function has a cylindrical symmetry. In general, the best practice is to use the coordinates that match the built-in symmetry of the function.

Integrals in Polar Coordinates

The Jacobian determinant of that transformation is the following:

[latex]\displaystyle{\frac{\partial (x,y)}{\partial (\rho, \phi)}} = \begin{vmatrix} \cos \phi & - \rho \sin \phi \\ \sin \phi & \rho \cos \phi \end{vmatrix} = \rho[/latex]

which has been obtained by inserting the partial derivatives of [latex]x = \rho \cos(\varphi)[/latex], [latex]y = \rho \sin(\varphi)[/latex] in the first column with respect to [latex]\rho[/latex] and in the second column with respect to [latex]\varphi[/latex], so the [latex]dx \, dy[/latex] differentials in this transformation become [latex]\rho \,d \rho \,d\varphi[/latex]. Once the function is transformed and the domain evaluated, it is possible to define the formula for the change of variables in polar coordinates:

[latex]\iint_D f(x,y)dx \, dy = \iint_T f(\rho \cos \varphi, \rho \sin \varphi)\rho[/latex]

Example

Integrate the function [latex]f(x,y) = x[/latex] over the domain:

[latex]D = \{ x^2 + y^2 \le 9, \ x^2 + y^2 \ge 4, \ y \ge 0 \}[/latex]

From [latex]f(x,y) = x \longrightarrow f(\rho,\phi) = \rho \cos \phi[/latex],

[latex]\begin{align} \iint_D x \, dx\, dy &= \iint_T \rho \cos \phi \rho \, d\rho\, d\phi \\ &= \int_0^\pi \int_2^3 \rho^2 \cos \phi \, d \rho \, d \phi \\ &= \int_0^\pi \cos \phi \ d \phi \left[ \frac{\rho^3}{3} \right]_2^3 \\ &= \left[ \sin \phi \right]_0^\pi \ \left(9 - \frac{8}{3} \right) = 0 \end{align}[/latex]

Triple Integrals in Cylindrical Coordinates

When the function to be integrated has a cylindrical symmetry, it is sensible to integrate using cylindrical coordinates.

Learning Objectives

Evaluate triple integrals in cylindrical coordinates

Key Takeaways

Key Points

  • Switching from Cartesian to cylindrical coordinates, the transformation of the function is made by the following relation [latex]f(x,y,z) \rightarrow f(\rho \cos \varphi, \rho \sin \varphi, z)[/latex].
  • In switching to cylindrical coordinates, the [latex]dx\, dy\, dz[/latex]  differentials in the integral become [latex]\rho \, d\rho \,d\varphi \,dz[/latex].
  • Therefore, an integral evaluated in Cartesian coordinates can be switched to an integral in cylindrical coordinates as[latex]\iiint_D f(x,y,z) \, dx\, dy\, dz = \iiint_T f(\rho \cos \varphi, \rho \sin \varphi, z)\rho \, d\rho \,d\varphi \,dz[/latex].

Key Terms

  • differential: an infinitesimal change in a variable, or the result of differentiation
  • cylindrical coordinate: a three-dimensional coordinate system that specifies point positions by the distance from a chosen reference axis, the direction from the axis relative to a chosen reference direction, and the distance from a chosen reference plane perpendicular to the axis

When the function to be integrated has a cylindrical symmetry, it is sensible to change the variables into cylindrical coordinates and then perform integration.

In R3 the integration on domains with a circular base can be made by the passage in cylindrical coordinates; the transformation of the function is made by the following relation:

[latex]f(x,y,z) \rightarrow f(\rho \cos \varphi, \rho \sin \varphi, z)[/latex]

The domain transformation can be graphically attained, because only the shape of the base varies, while the height follows the shape of the starting region. Also in switching to cylindrical coordinates, the [latex]dx\, dy\, dz[/latex] differentials in the integral become [latex]\rho \, d\rho \,d\varphi \,dz[/latex].

image

Cylindrical Coordinates: Cylindrical coordinates are often used for integrations on domains with a circular base.

Example 1

The region is:

[latex]D = \{ x^2 + y^2 \le 9, \ x^2 + y^2 \ge 4, \ 0 \le z \le 5 \}[/latex]

If the transformation is applied, this region is obtained:

[latex]T = \{ 2 \le \rho \le 3, \ 0 \le \varphi \le 2\pi, \ 0 \le z \le 5 \}[/latex]

because the z component is unvaried during the transformation, the [latex]dx\, dy\, dz[/latex] differentials vary as in the passage in polar coordinates: therefore, they become: [latex]\rho \, d\rho \,d\varphi \,dz[/latex]. Finally, it is possible to apply the final formula to cylindrical coordinates:

[latex]\displaystyle{\iiint Df(x,y,z)dx\,dy\,dz=\iiint Tf(\rho\cos\varphi,\rho\sin\varphi,z)\rho\, d\rho \,d\varphi \,dz}[/latex]

This method is convenient in case of cylindrical or conical domains or in regions where it is easy to individuate the [latex]z[/latex] interval and even transform the circular base and the function.

Example 2

The function [latex]f(x,y,z) = x^2 + y^2 + z[/latex] is and as integration domain this cylinder:

[latex]D = \{ x^2 + y^2 \le 9, \ -5 \le z \le 5 \}[/latex]

The transformation of [latex]D[/latex] in cylindrical coordinates is the following:

[latex]T = \{ 0 \le \rho \le 3, \ 0 \le \phi \le 2 \pi, \ -5 \le z \le 5 \}[/latex]

while the function becomes:

[latex]f(\rho \cos \varphi, \rho \sin \varphi, z) = \rho^2 + z[/latex]

Therefore, the integral becomes:

[latex]\begin{align}\displaystyle{\iiint_D (x^2 + y^2 +z) \, dx\, dy\, dz} &\displaystyle{= \iiint_T ( \rho^2 + z) \rho \, d\rho\, d\phi\, dz} \\ &= \int_{-5}^5 dz \int_0^{2 \pi} d\phi \int_0^3 ( \rho^3 + \rho z )\, d\rho\\ &= 2 \pi \int_{-5}^5 \left[ \frac{\rho^4}{4} + \frac{\rho^2 z}{2} \right]_0^3 \, dz\\ &= 2 \pi \int_{-5}^5 \left( \frac{81}{4} + \frac{9}{2} z\right)\, dz\\ &= 405 \pi \end{align}[/latex]

Triple Integrals in Spherical Coordinates

When the function to be integrated has a spherical symmetry, change the variables into spherical coordinates and then perform integration.

Learning Objectives

Evaluate triple integrals in spherical coordinates

Key Takeaways

Key Points

  • Switching from Cartesian to spherical coordinates, the function is transformed by this relation: [latex]f(x,y,z) \longrightarrow f(\rho \cos \theta \sin \varphi, \rho \sin \theta \sin \varphi, \rho \cos \varphi)[/latex].
  • For the transformation, the [latex]dx\, dy\, dz[/latex] differentials in the integral are transformed to [latex]\rho^2 \sin \varphi \, d\rho \,d\varphi \,dz[/latex].
  • Therefore, an integral evaluated in Cartesian coordinates can be switched to an integral in spherical coordinates as [latex]\iiint_D f(x,y,z) \, dx\, dy\, dz = \iiint_T f(\rho \sin \varphi \cos \theta, \rho \sin \varphi \sin \theta, \rho \cos \varphi) \rho^2 \sin \varphi \, d\rho\, d\theta\, d\varphi.[/latex]

Key Terms

  • spherical coordinate: a coordinate system for three-dimensional space where the position of a point is specified by three numbers: the radial distance of that point from a fixed origin, its polar angle measured from a fixed zenith direction, and the azimuth angle of its orthogonal projection on a reference plane that passes through the origin and is orthogonal to the zenith
  • Jacobian determinant: the determinant of the Jacobian matrix

When the function to be integrated has a spherical symmetry, it is sensible to change the variables into spherical coordinates and then perform integration.

image

Spherical Coordinates: Spherical coordinates are useful when domains in [latex]R^3[/latex] have spherical symmetry.

In [latex]R^3[/latex] some domains have a spherical symmetry, so it’s possible to specify the coordinates of every point of the integration region by two angles and one distance. It’s possible to use therefore the passage in spherical coordinates; the function is transformed by this relation:

[latex]f(x,y,z) \longrightarrow f(\rho \cos \theta \sin \varphi, \rho \sin \theta \sin \varphi, \rho \cos \varphi)[/latex]

Points on [latex]z[/latex]-axis do not have a precise characterization in spherical coordinates, so [latex]\theta[/latex] can vary from [latex]0[/latex] to [latex]2 \pi[/latex]. The better integration domain for this passage is obviously the sphere.

Integrals in Spherical Coordinates

The Jacobian determinant of this transformation is the following:

[latex]\displaystyle{\frac{\partial (x,y,z)}{\partial (\rho, \theta, \varphi)}} = \begin{vmatrix} \cos \theta \sin \varphi & - \rho \sin \theta \sin \varphi & \rho \cos \theta \cos \varphi \\ \sin \theta \sin \varphi & \rho \cos \theta \sin \varphi & \rho \sin \theta \cos \varphi \\ \cos \varphi & 0 & - \rho \sin \varphi \end{vmatrix} = \rho^2 \sin \varphi[/latex]

The [latex]dx\, dy\, dz[/latex] differentials therefore are transformed to [latex]\rho^2 \sin \varphi \, d\rho \,d\varphi \,dz[/latex].

Finally, you obtain the final integration formula: It’s better to use this method in case of spherical domains and in case of functions that can be easily simplified, by the first fundamental relation of trigonometry, extended in [latex]R^3[/latex]; in other cases it can be better to use cylindrical coordinates.

Example

Integrate [latex]f(x,y,z) = x^2 + y^2 + z^2[/latex] over the domain [latex]D = x^2 + y^2 + z^2 \le 16[/latex].

In spherical coordinates:

[latex]f(\rho \sin \varphi \cos \theta, \rho \sin \varphi \sin \theta, \rho \cos \varphi) = \rho^2[/latex]

while the intervals of the transformed region [latex]T[/latex] from [latex]D[/latex]:

[latex]0 \leq \rho \leq 4, 0 \leq \varphi \leq \pi, 0 \leq \theta \leq 2\pi[/latex]

Therefore:

[latex]\begin{align} \iiint_D (x^2 + y^2 +z^2) \, dx\, dy\, dz &= \iiint_T \rho^2 \ \rho^2 \sin \theta \, d\rho\, d\theta\, d\phi, \\ &= \int_0^{\pi} \sin \phi \,d\phi \int_0^4 \rho^4 d \rho \int_0^{2 \pi} d\theta \\ &= 2 \pi \int_0^{\pi} \sin \phi \left[ \frac{\rho^5}{5} \right]_0^4 \, d \phi \\ &= 2 \pi \left[ \frac{\rho^5}{5} \right]_0^4 \left[- \cos \phi \right]_0^{\pi}= \frac{4096 \pi}{5} \end{align}[/latex]

Triple Integrals

For [latex]T \subseteq R^3[/latex], the triple integral over [latex]T[/latex] is written as [latex]\iiint_T f(x,y,z)\, dx\, dy\, dz[/latex].

Learning Objectives

Use triple integrals to integrate over three-dimensional regions

Key Takeaways

Key Points

  • By convention, the triple integral has three integral signs (and a double integral has two integral signs); this is a notational convention which is convenient when computing a multiple integral as an iterated integral.
  • If [latex]T[/latex] is a domain that is normal with respect to the [latex]xy[/latex]-plane and determined by the functions [latex]\alpha (x,y)[/latex] and [latex]\beta(x,y)[/latex], then [latex]\iiint_T f(x,y,z) \ dx\, dy\, dz = \iint_D \int_{\alpha (x,y)}^{\beta (x,y)} f(x,y,z) \, dz dx dy[/latex].
  • To integrate a function with spherical symmetry such as [latex]f(x,y,z) = x^2 + y^2 + z^2[/latex], consider changing integration variable to spherical coordinates.

Key Terms

  • spherical coordinate: a coordinate system for three-dimensional space where the position of a point is specified by three numbers: the radial distance of that point from a fixed origin, its polar angle measured from a fixed zenith direction, and the azimuth angle of its orthogonal projection on a reference plane that passes through the origin and is orthogonal to the zenith

For [latex]T \subseteq R^3[/latex], the triple integral over [latex]T[/latex] is written as

[latex]\displaystyle{\iiint_T f(x,y,z)\, dx\, dy\, dz}[/latex]

Notice that, by convention, the triple integral has three integral signs (and a double integral has two integral signs); this is a notational convention which is convenient when computing a multiple integral as an iterated integral.

We have seen that double integrals can be evaluated over regions with a general shape. The extension of those formulae to triple integrals should be apparent. If [latex]T[/latex] is a domain that is normal with respect to the xy-plane and determined by the functions [latex]\alpha (x,y)[/latex] and [latex]\beta(x,y)[/latex], then:

[latex]\displaystyle{\iiint_T f(x,y,z) \ dx\, dy\, dz = \iint_D \int_{\alpha (x,y)}^{\beta (x,y)} f(x,y,z) \, dz dx dy}[/latex]

image

Graphical Representation of a Triple Integral: Example of domain in [latex]R^3[/latex] that is normal with respect to the [latex]xy[/latex]-plane.

Example 1

The volume of the parallelepiped of sides 4 by 6 by 5 may be obtained in two ways:

  • By calculating the double integral of the function [latex]f(x, y) = 5[/latex] over the region [latex]D[/latex] in the [latex]xy[/latex]-plane which is the base of the parallelepiped: [latex]\iint_D 5 \ dx\, dy[/latex]
  • By calculating the triple integral of the constant function 1 over the parallelepiped itself: [latex]\iiint_\mathrm{parallelepiped} 1 \, dx\, dy\, dz[/latex]

Example 2

Integrate [latex]f(x,y,z) = x^2 + y^2 + z^2[/latex] over the domain [latex]D = \left \{ x^2 + y^2 + z^2 \le 16 \right \}[/latex].

Looking at the domain, it seems convenient to adopt the passage in spherical coordinates; in fact, the intervals of the variables that delimit the new [latex]T[/latex] region are obviously:

[latex](0 \le \rho \le 4, \ 0 \le \phi \le \pi, \ 0 \le \theta \le 2 \pi)[/latex]

For the function, we get:

[latex]f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) = \rho^2[/latex]

Therefore:

[latex]\begin{align} \iiint_D (x^2 + y^2 +z^2) \, dx\, dy\, dz &= \iiint_T \rho^2 \ \rho^2 \sin \theta \, d\rho\, d\theta\, d\phi \\ &= \int_0^{\pi} \sin \phi \,d\phi \int_0^4 \rho^4 d \rho \int_0^{2 \pi} d\theta \\ &= 2 \pi \int_0^{\pi} \sin \phi \left[ \frac{\rho^5}{5} \right]_0^4 \, d \phi \\ &= 2 \pi \left[ \frac{\rho^5}{5} \right]_0^4 \left[- \cos \phi \right]_0^{\pi} = \frac{4096 \pi}{5} \end{align}[/latex]

Change of Variables

One makes a change of variables to rewrite the integral in a more “comfortable” region, which can be described in simpler formulae.

Learning Objectives

Use a change a variables to rewrite an integral in a more familiar region

Key Takeaways

Key Points

  • There exist three main “kinds” of changes of variable (to polar coordinate in [latex]R^2[/latex], and to cylindrical and spherical coordinates in [latex]R^3[/latex]); however, more general substitutions can be made using the same principle.
  • When changing integration variables, make sure that the integral domain also changes accordingly.
  • Change of variable should be judiciously applied based on the built-in symmetry of the function to be integrated.

Key Terms

  • polar coordinate: a two-dimensional coordinate system in which each point on a plane is determined by a distance from a fixed point and an angle from a fixed direction

The limits of integration are often not easily interchangeable (without normality or with complex formulae to integrate). One makes a change of variables to rewrite the integral in a more “comfortable” region, which can be described in simpler formulae. To do so, the function must be adapted to the new coordinates.

For example, for the function [latex]f(x, y) = (x-1)^2 +\sqrt y[/latex], if one adopts this substitution [latex]x' = x-1, \ y'= y[/latex], therefore [latex]x = x' + 1, \ y=y'[/latex], one obtains the new function:

[latex]f_2(x,y) = (x')^2 +\sqrt y[/latex]

which is simpler than the original form. When changing integration variables, however, make sure that the integral domain also changes accordingly. In the example, if integration is performed over [latex]x[/latex] in [latex][0,1][/latex]] and y in [latex][0,3][/latex], the new variables [latex]x'[/latex] and [latex]y'[/latex] vary over [latex][-1,0][/latex] and [latex][0,3][/latex], respectively. There exist three main “kinds” of changes of variable (one in [latex]R^2[/latex], two in [latex]R^3[/latex]); however, more general substitutions can be made using the same principle.

1. Polar coordinates

The function to be integrated transforms as:

[latex]f(x,y) \rightarrow f(\rho \cos \phi,\rho \sin \phi )[/latex]

and the integral accordingly changes as:

[latex]\displaystyle {\iint_D f(x,y) \ dx\, dy = \iint_T f(\rho \cos \phi, \rho \sin \phi) \rho \, d \rho\, d \phi}[/latex]

2. Cylindrical coordinates

The function to be integrated transforms as:

[latex]f(x,y,z) \rightarrow f(\rho \cos \phi, \rho \sin \phi, z)[/latex]

and the integral accordingly changes as:

[latex]\displaystyle {\iiint_D f(x,y,z) \, dx\, dy\, dz = \iiint_T f(\rho \cos \phi, \rho \sin \phi, z) \rho \, d\rho\, d\phi\, dz}[/latex]

image

Cylindrical Coordinates: Changing to cylindrical coordinates may be useful depending on the setup of problem.

3. Spherical coordinates

The function to be integrated transforms as:

[latex]f(x,y,z) \longrightarrow f(\rho \cos \theta \sin \phi, \rho \sin \theta \sin \phi, \rho \cos \phi)[/latex]

and the integral accordingly changes as:

[latex]\displaystyle {\iiint_D f(x,y,z) \, dx\, dy\, dz \\ = \iiint_T f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^2 \sin \phi \, d\rho\, d\theta\, d\phi}[/latex]

Applications of Multiple Integrals

Multiple integrals are used in many applications in physics and engineering.

Learning Objectives

Apply multiple integrals to real world examples

Key Takeaways

Key Points

  • Given a set [latex]D \subseteq R^n[/latex] and an integrable function [latex]f[/latex] over [latex]D[/latex], the average value of [latex]f[/latex] over its domain is given by [latex]\bar{f} = \frac{1}{m(D)} \int_D f(x)dx[/latex], where [latex]m(D)[/latex] is the measure of [latex]D[/latex].
  • The gravitational potential associated with a mass distribution given by a mass measure [latex]dm[/latex] on three-dimensional Euclidean space [latex]R^3[/latex] is [latex]V(\mathbf{x}) = -\int_{\mathbf{R}^3} \frac{G}{|\mathbf{x} - \mathbf{r}|}\,dm(\mathbf{r})[/latex].
  • An electric field produced by a distribution of charges given by the volume charge density [latex]\rho (\vec r)[/latex] is obtained by a triple integral of a vector function: [latex]\vec E = \frac {1}{4 \pi \epsilon_0} \iiint \frac {\vec r - \vec r'}{\| \vec r - \vec r' \|^3} \rho (\vec r')\, {d}^3 r'[/latex].

Key Terms

  • Maxwell’s equations: a set of partial differential equations that, together with the Lorentz force law, form the foundation of classical electrodynamics, classical optics, and electric circuits
  • moment of inertia: a measure of a body’s resistance to a change in its angular rotation velocity

As is the case with one variable, one can use the multiple integral to find the average of a function over a given set. Given a set [latex]D \subseteq R^n[/latex] and an integrable function [latex]f[/latex] over [latex]D[/latex], the average value of [latex]f[/latex] over its domain is given by:

[latex]\displaystyle{\bar{f} = \frac{1}{m(D)} \int_D f(x)\, dx}[/latex]

where [latex]m(D)[/latex] is the measure of [latex]D[/latex]. Additionally, multiple integrals are used in many applications in physics and engineering. The examples below also show some variations in the notation.

Example 1

In mechanics, the moment of inertia is calculated as the volume integral (triple integral) of the density weighed with the square of the distance from the axis:

[latex]\displaystyle{I_z = \iiint_V \rho r^2\, dV}[/latex]

Example 2

The gravitational potential associated with a mass distribution given by a mass measure [latex]dm[/latex] on three-dimensional Euclidean space [latex]R^3[/latex] is:

[latex]\displaystyle{V(\mathbf{x}) = -\int_{\mathbf{R}^3} \frac{G}{\left|\mathbf{x} - \mathbf{r}\right|}\,dm(\mathbf{r})}[/latex]

If there is a continuous function [latex]\rho(x)[/latex] representing the density of the distribution at [latex]x[/latex], so that [latex]dm(x) = \rho (x)d^3x[/latex], where [latex]d^3x[/latex] is the Euclidean volume element, then the gravitational potential is:

[latex]\displaystyle{V(\mathbf{x}) = -\int_{\mathbf{R}^3} \frac{G}{\left|\mathbf{x}-\mathbf{r}\right|}\,\rho(\mathbf{r})\,d^3\mathbf{r}}[/latex]

image

A Mass to be Integrated: Points [latex]\mathbf{x}[/latex] and [latex]\mathbf{r}[/latex], with [latex]\mathbf{r}[/latex] contained in the distributed mass (gray) and differential mass [latex]dm(\mathbf{r})[/latex]  located at the point [latex]\mathbf{r}[/latex].

Example 3

In electromagnetism, Maxwell’s equations can be written using multiple integrals to calculate the total magnetic and electric fields. In the following example, the electric field produced by a distribution of charges given by the volume charge density [latex]\rho (\vec r)[/latex] is obtained by a triple integral of a vector function:

[latex]\displaystyle{\vec E = \frac {1}{4 \pi \epsilon_0} \iiint \frac {\vec r - \vec r'}{\| \vec r - \vec r' \|^3} \rho (\vec r')\, {d}^3 r'}[/latex]

This can also be written as an integral with respect to a signed measure representing the charge distribution.

Center of Mass and Inertia

The center of mass for a rigid body can be expressed as a triple integral.

Learning Objectives

Use multiple integrals to find the center of mass of a distribution of mass

Key Takeaways

Key Points

  • In the case of a single rigid body, the center of mass is fixed in relation to the body, and if the body has uniform density, it will be located at the centroid.
  • In the case of a system of particles [latex]P_i, i = 1, \cdots, n[/latex], each with mass [latex]m_i[/latex] that are located in space with coordinates [latex]\mathbf{r}_i, i = 1, \cdots, n[/latex], the coordinates [latex]\mathbf{R}[/latex] of the center of mass is given as [latex]\mathbf{R} = \frac{1}{M} \sum_{i=1}^n m_i \mathbf{r}_i[/latex].
  • If the mass distribution is continuous with the density [latex]\rho (r)[/latex] within a volume [latex]V[/latex], the center of mass is expressed as [latex]\mathbf R = \frac 1M \int_V\rho(\mathbf{r}) \mathbf{r} dV[/latex].

Key Terms

  • rigid body: an idealized solid whose size and shape are fixed and remain unaltered when forces are applied; used in Newtonian mechanics to model real objects
  • centroid: the point at the center of any shape, sometimes called the center of area or the center of volume

Multiple integrals are used in many applications in physics and engineering. In this atom, we will see how center of mass can be calculated using multiple integrals.

The center of mass is the unique point at the center of a distribution of mass in space that has the property that the weighted position vectors relative to this point sum to zero. In the case of a single rigid body, the center of mass is fixed in relation to the body, and if the body has uniform density, it will be located at the centroid. The center of mass may be located outside the physical body, as is sometimes the case for hollow or open-shaped objects, such as a horseshoe. In the case of a distribution of separate bodies, such as the planets of the Solar System, the center of mass may not correspond to the position of any individual member of the system.

image

Center of Mass: Two bodies orbiting around the center of mass inside one body

A System of Particles

In the case of a system of particles [latex]P_i, i = 1, \cdots, n[/latex], each with mass [latex]m_i[/latex] that are located in space with coordinates [latex]\mathbf{r}_i, i = 1, \cdots, n[/latex], the coordinates [latex]\mathbf{R}[/latex] of the center of mass satisfy the condition:

[latex]\displaystyle{\sum_{i=1}^n m_i(\mathbf{r}_i - \mathbf{R}) = 0}[/latex]

Solve this equation for [latex]\mathbf{R}[/latex] to obtain the formula:

[latex]\displaystyle{\mathbf{R} = \frac{1}{M} \sum_{i=1}^n m_i \mathbf{r}_i}[/latex]

where [latex]M[/latex] is the sum of the masses of all of the particles.

A Continuous Volume

If the mass distribution is continuous with the density [latex]\rho (r)[/latex] within a volume [latex]V[/latex], then the integral of the weighted position coordinates of the points in this volume relative to the center of mass [latex]\mathbf{R}[/latex] is zero; that is:

[latex]\displaystyle{\int_V \rho(\mathbf{r})(\mathbf{r}-\mathbf{R})dV = 0}[/latex]

Solve this equation for the coordinates [latex]\mathbf{R}[/latex] to obtain:

[latex]\displaystyle{\mathbf R = \frac 1M \int_V\rho(\mathbf{r}) \mathbf{r} dV}[/latex]

where [latex]M[/latex] is the total mass in the volume. The integral is over the three dimensional volume, so it is a triple integral.

Sours: https://courses.lumenlearning.com/boundless-calculus/chapter/multiple-integrals/

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Video transcript

Sours: https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions/double-integrals-topic/v/double-integral-1
  1. Echo park kits
  2. Cat leopard bearings
  3. Hylands medicine

Math Insight

Multivariable calculus includes six different generalizations of the familiar one-variable integral of a scalar-valued function over an interval. One can integrate functions over one-dimensional curves, two dimensional planar regions and surfaces, as well as three-dimensional volumes. When integrating over curves and surfaces, one can integral vector fields, where the one integrates either the tangential (for curves) or the normal (for surfaces) component of the vector field.

In this page, we outline the various integrals, methods you can use to solve them, and their relationship to the fundamental theorems.

The integrals covered in this page are:

Line integral of scalar-valued function

The line integral of a scalar-valued function $\dlsi(\vc{x})$ over a curve $\dlc$ is written as \begin{align*} \dslint. \end{align*} One physical interpretation of this line integral is that it gives the mass of a wire from its density $\dlsi$.

The only way we've encountered to evaluate this integral is the direct method. We must parametrize $\dlc$ by some function $\dllp(t)$, for $a \le t \le b$. Then, we can calculate the line integral by turning it into a regular one-variable integral of the form \begin{align*} \dslint = \dpslint. \end{align*}

Note that the “length” $d\als$ became $\| \dllp'(t)\|dt$. This quantity $\| \dllp'(t)\|$ measures how $\dllp(t)$ stretches or shrinks the interval $[a,b]$ as it maps it onto $\dlc$.

Line integral of a vector field

The line integral of a vector field $\dlvf(\vc{x})$ over a curve $\dlc$ is written as \begin{align*} \dlint. \end{align*} One physical interpretation of this line integral is the calculation of work when a particle moves along a path. If $\dlvf$ were a force acting on a particle moving along $\dlc$, then the integral would be the total work performed by the force on the particle.

This integral is one of the most important of multivariable calculus. We have four alternatives to evaluate the integral, although most of the alternatives work only in special cases.

  1. We can compute the integral directly. We parametrize $\dlc$ by some function $\dllp(t)$, for $a \le t \le b$. Then \begin{align*} \dlint = \dplint \end{align*}

    This method always applies. Sometimes, though, the integral will be difficult or we won't even be able to evaluate it. Our lives can be made easier by using one of the fundamental theorems to convert the line integral into something else.

    Since this integral is really a line integral of the scalar-valued function $\dlsi = \dlvf \cdot \vc{T}$ where $\vc{T}$ is the unit tangent vector \begin{align*} \vc{T} = \frac{ \dllp'(t)}{\| \dllp'(t)\|}, \end{align*} the formula for the direct method is the same as the formula for the scalar-valued path integral with $\dlsi = \dlvf \cdot \vc{T}$.

  2. If the vector field $\dlvf$ happens to be conservative, then we could use the gradient theorem for line integrals. We reduce the problem from an integral over the curve $\dlc$ to something just depending on the “boundary” of $\dlc$, i.e., its endpoints.

    We need to find a potential function $f$ so that $\nabla f = \dlvf$. Then, \begin{align*} \dlint = f(\vc{q}) - f(\vc{p}), \end{align*} where $\vc{p}$ and $\vc{q}$ are the endpoints of $\dlc$.

    If $\dlc$ also happens to be a closed curve, then the integral of a conservative vector field $\dlvf$ will be zero. Also, that if you know $\dlvf$ is conservative, another thing you can do is just change the curve $\dlc$ to another curve that has the same endpoints as $\dlc$. In this case, the line integral of $\dlvf$ over $\dlc$ is the same as the line integral of $\dlvf$ over any other curve with the same endpoints.

  3. If the vector field $\dlvf$ and the curve $\dlc$ happen to be in two dimensions and if $\dlc$ happens to be a simpleclosed curve, then we can use Green's theorem. Green's theorem converts the line integral over $\dlc$ to a double integral over the interior of $\dlc$, which we call $\dlr$, \begin{align*} \dlint = \iint_\dlr \left(\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y} \right) \, dA. \end{align*} Note that $\dlvf$ must be continuously differentiable everywhere in $\dlr$ for this to work. Sometimes we write $\dlc= \partial \dlr$ to denote that $\dlc$ is the boundary of $\dlr$. $\dlc$ must be oriented in a counterclockwise fashion, otherwise, we'll be off by a minus sign.

  4. If the vector field $\dlvf$ and the curve $\dlc$ happen to be in three dimensions and if $\dlc$ happens to be a simpleclosed curve, then we can use Stokes' theorem. Stokes' theorem converts the line integral over $\dlc$ to a surface integral over any surface $\dls$ for which $\dlc$ is a boundary, \begin{align*} \dlint = \sint{\dls}{\curl \dlvf}, \end{align*} and is valid for any surface over which $\dlvf$ is continuously differentiable. Sometimes we write $\dlc= \partial \dls$ to denote that $\dlc$ is the boundary of $\dls$. $\dlc$ must be a positively oriented boundary of $\dls$, otherwise, we'll be off by a minus sign.

Surface integral of a scalar-valued function

The surface integral of a scalar-valued function $\dlsi(\vc{x})$ over a surface $\dls$ is written as \begin{align*} \dssint \end{align*} One physical interpretation of the integral is the mass of a sheet with density given by $\dlsi$.

The only way we've encountered to evalute this integral is the direct method. We must parametrize $\dls$ by some function $\dlsp(\spfv,\spsv)$, for $(\spfv,\spsv) \in \dlr$. Then, \begin{align*} \dssint = \dpssint \end{align*}

Note that the “area” $d\sas$ became $\left\| \pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right\| \,d\spfv\,d\spsv$. The quantity $d\sas$ became $\left\| \pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right\|$ measures how $\dlsp(\spfv,\spsv)$ stretches or shrinks the region $\dlr$ as it maps it onto $\dls$.

Surface integral of a vector field

The surface integral over surface $\dls$ of a vector field $\dlvf(\vc{x})$ is written as \begin{align*} \dsint. \end{align*} A physical interpretation is the flux of a fluid through $\dls$ whose velocity is given by $\dlvf$. For this reason, we sometimes refer to the integral as a “flux integral.”

Like the line integral of a vector field, this integral plays a big role in this course. We have three alternatives to evaluate the integral, although most of the alternatives work only in special cases.

  1. We can compute the integral directly. We parametrize $\dls$ by some function $\dlsp(\spfv,\spsv)$, for $(\spfv,\spsv) \in \dlr$. Then, \begin{align*} \dsint = \dpsint. \end{align*}

    This method always applies. Sometimes, though, the integral will be difficult or we won't even be able to evaluate it. Our lives can be made easier by using one of the fundamental theorems to convert the surface integral into something else.

    Since this integral is really a surface integral of the scalar-valued function $\dlsi = \dlvf \cdot \vc{n}$ where $\vc{n}$ is the unit normal vector \begin{align*} \vc{n} = \frac{\displaystyle\pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}} {\displaystyle\left\|\pdiff{\dlsp}{\spfv} \times \pdiff{\dlsp}{\spsv}\right\|}, \end{align*} the formula for the direct method is the same as the formula for the scalar-valued surface integral with $\dlsi = \dlvf \cdot \vc{n}$.

  2. If the vector field $\dlvf$ happens to be the curl of another vector field $\vc{G}$, i.e., $\dlvf = \curl \vc{G}$, then we can apply Stokes' theorem to convert the surface integral of $\curl \vc{G}$ into the line integral of $\vc{G}$ around the positively oriented boundary of $\dls$, which we denote $\partial \dls$, \begin{align*} \dsint = \sint{\dls}{\curl \vc{G}} =\lint{\dlc}{\vc{G}} \end{align*}

    Multivariable calculus student's don't typically learn methods to find $\vc{G}$ from $\dlvf$. We can use Stokes' theorem to convert a surface integral into a line integral only if we are told outright that $\dlvf = \curl \vc{G}$ and are given what $\vc{G}$ is. But, if given the surface integral that looks like $\sint{\dls}{\curl \vc{G}}$, we can immediately recognize that Stokes' theorem is an option.

    Stokes' theorem allows us to do one more thing to the integral $\sint{\dls}{\curl \vc{G}}$. We can switch the surface $\dls$ to any other surface $\dls'$ as long as the boundaries of $\dls$ and $\dls'$ are the same, i.e., $\partial \dls = \partial \dls'$ (assuming both boundaries are positively oriented). If $\dls$ is a complicated surface, we could feasibly save ourselves some work by integrating over another surface $\dls'$ if that surface is simpler than $\dls$.

  3. If the surface $\dls$ happens to be a closed surface so that it is the boundary of some solid $\dlv$, i.e., $\dls = \partial \dlv$, then we can use the divergence theorem to convert the surface integral into the triple integral of $\div \dlvf$ over $\dlv$, \begin{align*} \dsint = \iiint_\dlv \div \dlvf \, dV, \end{align*} where we orient $\dls$ so that it has an outward pointing normal vector. This works only if $\dlvf$ is continuously differentiable everywhere in the solid $\dlv$.

Double integrals

The double integral of a (scalar-valued) function $\dlsi(\vc{x})$ over a two-dimensional region $\dlr$ is written as \begin{align*} \iint_\dlr \dlsi \, dA. \end{align*} One physical interpretation is the mass of a region with density \dlsi$.

We have encountered three alternatives to evaluate the integral.

  1. We can compute the integral directly in terms of the original variables $x$ and $y$. In this case, $dA = dx\,dy$.

  2. We can compute the integral by changing to the variables $\cvarfv$ and $\cvarsv$ by finding a function $(x,y) = \vc{T}(\cvarfv,\cvarsv)$. Then the integral is \begin{align*} \iint_\dlr \dlsi\, dA = \left.\iint_{\dlr^{\textstyle *}}\right. \dlsi(\vc{T}(\cvarfv,\cvarsv)) \left| \det \jacm{\vc{T}}(\cvarfv,\cvarsv) \right| d\cvarfv\,d\cvarsv, \end{align*} where $\dlr$ is parametrized by $(x,y)=\vc{T}(\cvarfv,\cvarsv)$ for $(\cvarfv,\cvarsv)$ in $\dlr^{\textstyle *}$. We sometimes write the determinant of the matrix of partial derivatives of $\vc{T}(\cvarfv,\cvarsv)$ as $\displaystyle \det \jacm{\vc{T}}(\cvarfv,\cvarsv) = \pdiff{(x,y)}{(\cvarfv,\cvarsv)}$.

  3. If $f$ happens to be equal to $\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}$ for some vector field $\dlvf$, then we could use Green's theorem to convert the double integral into the integral of $\dlvf$ around the boundary of $\dlr$, which we denote $\partial \dlr$, \begin{align*} \iint_\dlr f \, dA = \iint_\dlr \left(\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}\right) dA = \lint{\partial \dlr}{\dlvf}. \end{align*} To orient the boundary properly, outside boundaries must be counterclockwise and inner boundaries must be clockwise.

    We usually think of Green's theorem going the other way, i.e., converting a line integral into a double integral. One reason for this is that multivariable calculus students don't usually learn methods to find $\dlvf$ from $f$. We can use Green's theorem to convert a double integral into a line integral only if we are told outright that $f =\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}$ and are given what $\dlvf$ is. But, if given the double integral that looks like $\iint_\dlr \left(\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}\right) dA$, we can immediately recognize that Green's theorem is an option.

    As a special case, if we are given an integral $\iint_\dlr dA$ (i.e., finding the area), we can let $\dlvf(x,y) = (-y,x)/2$ so that $\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y} =1$ and $\iint_\dlr dA = \lint{\partial \dlr}{\dlvf}$.

Triple integrals

The triple integral of a (scalar-valued) function $\dlsi(\vc{x})$ over a three-dimensional solid $\dlv$ is written as \begin{align*} \iiint_\dlv \dlsi \, dV. \end{align*} One physical interpretation is the mass of a solid with density given by $\dlsi$.

We have encountered three alternatives to evaluate the integral.

  1. We can compute the integral directly in terms of the original variables $x$, $y$, and $z$. In this case, $dV = dx\,dy\,dz$.

  2. We can compute the integral by changing to the variables $\cvarfv$, $\cvarsv$, and $\cvartv$ by finding a function $(x,y,z) = \vc{T}(\cvarfv,\cvarsv,\cvartv)$. Then the integral is \begin{align*} \iiint_\dlv \dlsi\, dV = \left.\iiint_{\dlv^{\textstyle *}}\right. \dlsi(\vc{T}(\cvarfv,\cvarsv,\cvartv)) \left| \det \jacm{\vc{T}}(\cvarfv,\cvarsv,\cvartv) \right| d\cvarfv\,d\cvarsv\,d\cvartv, \end{align*} where $\dlv$ is parametrized by $(x,y,z)=\vc{T}(\cvarfv,\cvarsv,\cvartv)$ for $(\cvarfv,\cvarsv,\cvartv)$ in $\dlv^{\textstyle *}$. We sometimes write the determinant of the matrix of partial derivatives of $\vc{T}(\cvarfv,\cvarsv,\cvartv)$ as $\displaystyle \det \jacm{\vc{T}}(\cvarfv,\cvarsv,\cvartv) = \pdiff{(x,y,z)}{(\cvarfv,\cvarsv,\cvartv)}$.

  3. If $\dlsi$ happens to be equal to $\div \dlvf$ for some vector field $\dlvf$, then we could use the divergence theorem to convert the triple integral into the surface integral of $\dlvf$ around the boundary of $\dlv$, which we denote $\partial \dlv$, \begin{align*} \iiint_\dlv \dlsi\, dV = \iiint_\dlv \div \dlvf\, dV = \sint{\partial \dlv}{\dlvf}. \end{align*}

    We usually think of the divergence theorem going the other way, i.e., converting a surface integral into a triple integral. One reason for this is that multivariable calculus student don't typically learn methods to find $\dlvf$ from $f$. We can use the divergence theorem to convert a triple integral into a surface integral only if we are told outright that $\dlsi = \div \dlvf$ and are given what $\dlvf$ is. But, if given a triple integral that looks like $\iiint_\dlv \div\dlvf \, dV$, we can immediately recognize that the divergence theorem is an option.

Sours: https://mathinsight.org/integrals_multivariable_calculus_summary
Double Integrals

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About this unit

There are many ways to extend the idea of integration to multiple dimensions: Line integrals, double integrals, triple integrals, surface integrals, etc. Each one lets you add infinitely many infinitely small values, where those values might come from points on a curve, points in an area, points on a surface, etc. These are all very powerful tools, relevant to almost all real-world applications of calculus. In particular, they are an invaluable tool in physics.
Sours: https://www.khanacademy.org/math/multivariable-calculus/integrating-multivariable-functions

Calculus integration multivariable

Multiple integral

Generalization of definite integrals to functions of multiple variables

Integral as area between two curves.
Double integral as volume under a surface z = 10 − x2y2/8. The rectangular region at the bottom of the body is the domain of integration, while the surface is the graph of the two-variable function to be integrated.

In mathematics (specifically multivariable calculus), a multiple integral is a definite integral of a function of several real variables, for instance, f(x, y) or f(x, y, z). Integrals of a function of two variables over a region in \mathbb {R} ^{2} (the real-number plane) are called double integrals, and integrals of a function of three variables over a region in \mathbb {R} ^{3} (real-number 3D space) are called triple integrals.[1] For multiple integrals of a single-variable function, see the Cauchy formula for repeated integration.

Introduction[edit]

Just as the definite integral of a positive function of one variable represents the area of the region between the graph of the function and the x-axis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function (on the three-dimensional Cartesian plane where z = f(x, y)) and the plane which contains its domain.[1] If there are more variables, a multiple integral will yield hypervolumes of multidimensional functions.

Multiple integration of a function in n variables: f(x1, x2, ..., xn) over a domain D is most commonly represented by nested integral signs in the reverse order of execution (the leftmost integral sign is computed last), followed by the function and integrand arguments in proper order (the integral with respect to the rightmost argument is computed last). The domain of integration is either represented symbolically for every argument over each integral sign, or is abbreviated by a variable at the rightmost integral sign:[2]

{\displaystyle \int \cdots \int _{\mathbf {D} }\,f(x_{1},x_{2},\ldots ,x_{n})\,dx_{1}\!\cdots dx_{n}}

Since the concept of an antiderivative is only defined for functions of a single real variable, the usual definition of the indefinite integral does not immediately extend to the multiple integral.

Mathematical definition[edit]

For n > 1, consider a so-called "half-open" n-dimensional hyperrectangular domain T, defined as:

{\displaystyle T=[a_{1},b_{1})\times [a_{2},b_{2})\times \cdots \times [a_{n},b_{n})\subseteq \mathbb {R} ^{n}.}

Partition each interval [aj, bj) into a finite family Ij of non-overlapping subintervals ijα, with each subinterval closed at the left end, and open at the right end.

Then the finite family of subrectangles C given by

C=I_{1}\times I_{2}\times \cdots \times I_{n}

is a partition of T; that is, the subrectangles Ck are non-overlapping and their union is T.

Let f : TR be a function defined on T. Consider a partition C of T as defined above, such that C is a family of m subrectangles Cm and

T=C_{1}\cup C_{2}\cup \cdots \cup C_{m}

We can approximate the total (n + 1)th-dimensional volume bounded below by the n-dimensional hyperrectangle T and above by the n-dimensional graph of f with the following Riemann sum:

\sum _{k=1}^{m}f(P_{k})\,\operatorname {m} (C_{k})

where Pk is a point in Ck and m(Ck) is the product of the lengths of the intervals whose Cartesian product is Ck, also known as the measure of Ck.

The diameter of a subrectangle Ck is the largest of the lengths of the intervals whose Cartesian product is Ck. The diameter of a given partition of T is defined as the largest of the diameters of the subrectangles in the partition. Intuitively, as the diameter of the partition C is restricted smaller and smaller, the number of subrectangles m gets larger, and the measure m(Ck) of each subrectangle grows smaller. The function f is said to be Riemann integrable if the limit

{\displaystyle S=\lim _{\delta \to 0}\sum _{k=1}^{m}f(P_{k})\,\operatorname {m} (C_{k})}

exists, where the limit is taken over all possible partitions of T of diameter at most δ.[3]

If f is Riemann integrable, S is called the Riemann integral of f over T and is denoted

{\displaystyle \int \cdots \int _{T}\,f(x_{1},x_{2},\ldots ,x_{n})\,dx_{1}\!\cdots dx_{n}}

Frequently this notation is abbreviated as

\int _{T}\!f(\mathbf {x} )\,d^{n}\mathbf {x} .

where x represents the n-tuple (x1, …, xn) and dnx is the n-dimensional volume differential.

The Riemann integral of a function defined over an arbitrary bounded n-dimensional set can be defined by extending that function to a function defined over a half-open rectangle whose values are zero outside the domain of the original function. Then the integral of the original function over the original domain is defined to be the integral of the extended function over its rectangular domain, if it exists.

In what follows the Riemann integral in n dimensions will be called the multiple integral.

Properties[edit]

Multiple integrals have many properties common to those of integrals of functions of one variable (linearity, commutativity, monotonicity, and so on). One important property of multiple integrals is that the value of an integral is independent of the order of integrands under certain conditions. This property is popularly known as Fubini's theorem.[4]

Particular cases[edit]

In the case of {\displaystyle T\subseteq \mathbb {R} ^{2}}, the integral

{\displaystyle l=\iint _{T}f(x,y)\,dx\,dy}

is the double integral of f on T, and if {\displaystyle T\subseteq \mathbb {R} ^{3}} the integral

{\displaystyle l=\iiint _{T}f(x,y,z)\,dx\,dy\,dz}

is the triple integral of f on T.

Notice that, by convention, the double integral has two integral signs, and the triple integral has three; this is a notational convention which is convenient when computing a multiple integral as an iterated integral, as shown later in this article.

Methods of integration[edit]

The resolution of problems with multiple integrals consists, in most cases, of finding a way to reduce the multiple integral to an iterated integral, a series of integrals of one variable, each being directly solvable. For continuous functions, this is justified by Fubini's theorem. Sometimes, it is possible to obtain the result of the integration by direct examination without any calculations.

The following are some simple methods of integration:[1]

Integrating constant functions[edit]

When the integrand is a constant functionc, the integral is equal to the product of c and the measure of the domain of integration. If c = 1 and the domain is a subregion of R2, the integral gives the area of the region, while if the domain is a subregion of R3, the integral gives the volume of the region.

Example. Let f(x, y) = 2 and

{\displaystyle D=\left\{(x,y)\in \mathbb {R} ^{2}\ :\ 2\leq x\leq 4\ ;\ 3\leq y\leq 6\right\}}

in which case

{\displaystyle \int _{3}^{6}\int _{2}^{4}\ 2\ dx\,dy=2\int _{3}^{6}\int _{2}^{4}\ 1\ dx\,dy=2\cdot \operatorname {area} (D)=2\cdot (2\cdot 3)=12,}

since by definition we have:

{\displaystyle \int _{3}^{6}\int _{2}^{4}\ 1\ dx\,dy=\operatorname {area} (D).}

Use of symmetry[edit]

When the domain of integration is symmetric about the origin with respect to at least one of the variables of integration and the integrand is odd with respect to this variable, the integral is equal to zero, as the integrals over the two halves of the domain have the same absolute value but opposite signs. When the integrand is even with respect to this variable, the integral is equal to twice the integral over one half of the domain, as the integrals over the two halves of the domain are equal.

Example 1. Consider the function f(x,y) = 2 sin(x) − 3y3 + 5 integrated over the domain

{\displaystyle T=\left\{(x,y)\in \mathbb {R} ^{2}\ :\ x^{2}+y^{2}\leq 1\right\},}

a disc with radius 1 centered at the origin with the boundary included.

Using the linearity property, the integral can be decomposed into three pieces:

{\displaystyle \iint _{T}\left(2\sin x-3y^{3}+5\right)\,dx\,dy=\iint _{T}2\sin x\,dx\,dy-\iint _{T}3y^{3}\,dx\,dy+\iint _{T}5\,dx\,dy}

The function 2 sin(x) is an odd function in the variable x and the disc T is symmetric with respect to the y-axis, so the value of the first integral is 0. Similarly, the function 3y3 is an odd function of y, and T is symmetric with respect to the x-axis, and so the only contribution to the final result is that of the third integral. Therefore the original integral is equal to the area of the disk times 5, or 5π.

Example 2. Consider the function f(x, y, z) = x exp(y2 + z2) and as integration region the ball with radius 2 centered at the origin,

{\displaystyle T=\left\{(x,y,z)\in \mathbb {R} ^{3}\ :\ x^{2}+y^{2}+z^{2}\leq 4\right\}.}

The "ball" is symmetric about all three axes, but it is sufficient to integrate with respect to x-axis to show that the integral is 0, because the function is an odd function of that variable.

Normal domains on R2[edit]

See also: Order of integration (calculus)

This method is applicable to any domain D for which:

  • the projection of D onto either the x-axis or the y-axis is bounded by the two values, a and b
  • any line perpendicular to this axis that passes between these two values intersects the domain in an interval whose endpoints are given by the graphs of two functions, α and β.

Such a domain will be here called a normal domain. Elsewhere in the literature, normal domains are sometimes called type I or type II domains, depending on which axis the domain is fibred over. In all cases, the function to be integrated must be Riemann integrable on the domain, which is true (for instance) if the function is continuous.

x-axis[edit]

If the domain D is normal with respect to the x-axis, and f : DR is a continuous function; then α(x) and β(x) (both of which are defined on the interval [a, b]) are the two functions that determine D. Then, by Fubini's theorem:[5]

{\displaystyle \iint _{D}f(x,y)\,dx\,dy=\int _{a}^{b}dx\int _{\alpha (x)}^{\beta (x)}f(x,y)\,dy.}

y-axis[edit]

If D is normal with respect to the y-axis and f : DR is a continuous function; then α(y) and β(y) (both of which are defined on the interval [a, b]) are the two functions that determine D. Again, by Fubini's theorem:

{\displaystyle \iint _{D}f(x,y)\,dx\,dy=\int _{a}^{b}dy\int _{\alpha (y)}^{\beta (y)}f(x,y)\,dx.}

Normal domains on R3[edit]

If T is a domain that is normal with respect to the xy-plane and determined by the functions α(x, y) and β(x, y), then

{\displaystyle \iiint _{T}f(x,y,z)\,dx\,dy\,dz=\iint _{D}\int _{\alpha (x,y)}^{\beta (x,y)}f(x,y,z)\,dz\,dx\,dy}

This definition is the same for the other five normality cases on R3. It can be generalized in a straightforward way to domains in Rn.

Change of variables[edit]

See also: Integration by substitution § Substitution for multiple variables

The limits of integration are often not easily interchangeable (without normality or with complex formulae to integrate). One makes a change of variables to rewrite the integral in a more "comfortable" region, which can be described in simpler formulae. To do so, the function must be adapted to the new coordinates.

Example 1a. The function is f(x, y) = (x − 1)2 + √y; if one adopts the substitution u = x − 1, v = y therefore x = u + 1, y = v one obtains the new function f2(u, v) = (u)2 + √v.

  • Similarly for the domain because it is delimited by the original variables that were transformed before (x and y in example).
  • the differentials dx and dy transform via the absolute value of the determinant of the Jacobian matrix containing the partial derivatives of the transformations regarding the new variable (consider, as an example, the differential transformation in polar coordinates).

There exist three main "kinds" of changes of variable (one in R2, two in R3); however, more general substitutions can be made using the same principle.

Polar coordinates[edit]

See also: Polar coordinate system

Transformation from cartesian to polar coordinates.

In R2 if the domain has a circular symmetry and the function has some particular characteristics one can apply the transformation to polar coordinates (see the example in the picture) which means that the generic points P(x, y) in Cartesian coordinates switch to their respective points in polar coordinates. That allows one to change the shape of the domain and simplify the operations.

The fundamental relation to make the transformation is the following:

{\displaystyle f(x,y)\rightarrow f(\rho \cos \varphi ,\rho \sin \varphi ).}

Example 2a. The function is f(x, y) = x + y and applying the transformation one obtains

{\displaystyle f(\rho ,\varphi )=\rho \cos \varphi +\rho \sin \varphi =\rho (\cos \varphi +\sin \varphi ).}

Example 2b. The function is f(x, y) = x2 + y2, in this case one has:

{\displaystyle f(\rho ,\varphi )=\rho ^{2}\left(\cos ^{2}\varphi +\sin ^{2}\varphi \right)=\rho ^{2}}

using the Pythagorean trigonometric identity (very useful to simplify this operation).

The transformation of the domain is made by defining the radius' crown length and the amplitude of the described angle to define the ρ, φ intervals starting from x, y.

Example of a domain transformation from cartesian to polar.

Example 2c. The domain is D = {x2 + y2 ≤ 4}, that is a circumference of radius 2; it's evident that the covered angle is the circle angle, so φ varies from 0 to 2π, while the crown radius varies from 0 to 2 (the crown with the inside radius null is just a circle).

Example 2d. The domain is D = {x2 + y2 ≤ 9, x2 + y2 ≥ 4, y ≥ 0}, that is the circular crown in the positive y half-plane (please see the picture in the example); φ describes a plane angle while ρ varies from 2 to 3. Therefore the transformed domain will be the following rectangle:

{\displaystyle T=\{2\leq \rho \leq 3,\ 0\leq \varphi \leq \pi \}.}

The Jacobian determinant of that transformation is the following:

{\displaystyle {\frac {\partial (x,y)}{\partial (\rho ,\varphi )}}={\begin{vmatrix}\cos \varphi &-\rho \sin \varphi \\\sin \varphi &\rho \cos \varphi \end{vmatrix}}=\rho }

which has been obtained by inserting the partial derivatives of x = ρ cos(φ), y = ρ sin(φ) in the first column respect to ρ and in the second respect to φ, so the dx dy differentials in this transformation become ρ dρ dφ.

Once the function is transformed and the domain evaluated, it is possible to define the formula for the change of variables in polar coordinates:

{\displaystyle \iint _{D}f(x,y)\,dx\,dy=\iint _{T}f(\rho \cos \varphi ,\rho \sin \varphi )\rho \,d\rho \,d\varphi .}

φ is valid in the [0, 2π] interval while ρ, which is a measure of a length, can only have positive values.

Example 2e. The function is f(x, y) = x and the domain is the same as in Example 2d. From the previous analysis of D we know the intervals of ρ (from 2 to 3) and of φ (from 0 to π). Now we change the function:

{\displaystyle f(x,y)=x\longrightarrow f(\rho ,\varphi )=\rho \cos \varphi .}

finally let's apply the integration formula:

{\displaystyle \iint _{D}x\,dx\,dy=\iint _{T}\rho \cos \varphi \rho \,d\rho \,d\varphi .}

Once the intervals are known, you have

{\displaystyle \int _{0}^{\pi }\int _{2}^{3}\rho ^{2}\cos \varphi \,d\rho \,d\varphi =\int _{0}^{\pi }\cos \varphi \ d\varphi \left[{\frac {\rho ^{3}}{3}}\right]_{2}^{3}={\Big [}\sin \varphi {\Big ]}_{0}^{\pi }\ \left(9-{\frac {8}{3}}\right)=0.}

Cylindrical coordinates[edit]

In R3 the integration on domains with a circular base can be made by the passage to cylindrical coordinates; the transformation of the function is made by the following relation:

{\displaystyle f(x,y,z)\rightarrow f(\rho \cos \varphi ,\rho \sin \varphi ,z)}

The domain transformation can be graphically attained, because only the shape of the base varies, while the height follows the shape of the starting region.

Example 3a. The region is D = {x2 + y2 ≤ 9, x2 + y2 ≥ 4, 0 ≤ z ≤ 5} (that is the "tube" whose base is the circular crown of Example 2d and whose height is 5); if the transformation is applied, this region is obtained:

{\displaystyle T=\{2\leq \rho \leq 3,\ 0\leq \varphi \leq 2\pi ,\ 0\leq z\leq 5\}}

(that is, the parallelepiped whose base is similar to the rectangle in Example 2d and whose height is 5).

Because the z component is unvaried during the transformation, the dx dy dz differentials vary as in the passage to polar coordinates: therefore, they become ρ dρ dφ dz.

Finally, it is possible to apply the final formula to cylindrical coordinates:

{\displaystyle \iiint _{D}f(x,y,z)\,dx\,dy\,dz=\iiint _{T}f(\rho \cos \varphi ,\rho \sin \varphi ,z)\rho \,d\rho \,d\varphi \,dz.}

This method is convenient in case of cylindrical or conical domains or in regions where it is easy to individuate the z interval and even transform the circular base and the function.

Example 3b. The function is f(x, y, z) = x2 + y2 + z and as integration domain this cylinder: D = {x2 + y2 ≤ 9, −5 ≤ z ≤ 5}. The transformation of D in cylindrical coordinates is the following:

{\displaystyle T=\{0\leq \rho \leq 3,\ 0\leq \varphi \leq 2\pi ,\ -5\leq z\leq 5\}.}

while the function becomes

{\displaystyle f(\rho \cos \varphi ,\rho \sin \varphi ,z)=\rho ^{2}+z}

Finally one can apply the integration formula:

{\displaystyle \iiint _{D}\left(x^{2}+y^{2}+z\right)\,dx\,dy\,dz=\iiint _{T}\left(\rho ^{2}+z\right)\rho \,d\rho \,d\varphi \,dz;}

developing the formula you have

{\displaystyle \int _{-5}^{5}dz\int _{0}^{2\pi }d\varphi \int _{0}^{3}\left(\rho ^{3}+\rho z\right)\,d\rho =2\pi \int _{-5}^{5}\left[{\frac {\rho ^{4}}{4}}+{\frac {\rho ^{2}z}{2}}\right]_{0}^{3}\,dz=2\pi \int _{-5}^{5}\left({\frac {81}{4}}+{\frac {9}{2}}z\right)\,dz=\cdots =405\pi .}

Spherical coordinates[edit]

In R3 some domains have a spherical symmetry, so it's possible to specify the coordinates of every point of the integration region by two angles and one distance. It's possible to use therefore the passage to spherical coordinates; the function is transformed by this relation:

{\displaystyle f(x,y,z)\longrightarrow f(\rho \cos \theta \sin \varphi ,\rho \sin \theta \sin \varphi ,\rho \cos \varphi )}

Points on the z-axis do not have a precise characterization in spherical coordinates, so θ can vary between 0 and 2π.

The better integration domain for this passage is the sphere.

Example 4a. The domain is D = x2 + y2 + z2 ≤ 16 (sphere with radius 4 and center at the origin); applying the transformation you get the region

{\displaystyle T=\{0\leq \rho \leq 4,\ 0\leq \varphi \leq \pi ,\ 0\leq \theta \leq 2\pi \}.}

The Jacobian determinant of this transformation is the following:

{\displaystyle {\frac {\partial (x,y,z)}{\partial (\rho ,\theta ,\varphi )}}={\begin{vmatrix}\cos \theta \sin \varphi &-\rho \sin \theta \sin \varphi &\rho \cos \theta \cos \varphi \\\sin \theta \sin \varphi &\rho \cos \theta \sin \varphi &\rho \sin \theta \cos \varphi \\\cos \varphi &0&-\rho \sin \varphi \end{vmatrix}}=\rho ^{2}\sin \varphi }

The dx dy dz differentials therefore are transformed to ρ2 sin(φ) .

This yields the final integration formula:

{\displaystyle \iiint _{D}f(x,y,z)\,dx\,dy\,dz=\iiint _{T}f(\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi )\rho ^{2}\sin \varphi \,d\rho \,d\theta \,d\varphi .}

It is better to use this method in case of spherical domains and in case of functions that can be easily simplified by the first fundamental relation of trigonometry extended to R3 (see Example 4b); in other cases it can be better to use cylindrical coordinates (see Example 4c).

{\displaystyle \iiint _{T}f(a,b,c)\rho ^{2}\sin \varphi \,d\rho \,d\theta \,d\varphi .}

The extra ρ2 and sin φ come from the Jacobian.

In the following examples the roles of φ and θ have been reversed.

Example 4b.D is the same region as in Example 4a and f(x, y, z) = x2 + y2 + z2 is the function to integrate. Its transformation is very easy:

{\displaystyle f(\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi )=\rho ^{2},}

while we know the intervals of the transformed region T from D:

{\displaystyle T=\{0\leq \rho \leq 4,\ 0\leq \varphi \leq \pi ,\ 0\leq \theta \leq 2\pi \}.}

We therefore apply the integration formula:

{\displaystyle \iiint _{D}\left(x^{2}+y^{2}+z^{2}\right)\,dx\,dy\,dz=\iiint _{T}\rho ^{2}\,\rho ^{2}\sin \theta \,d\rho \,d\theta \,d\varphi ,}

and, developing, we get

{\displaystyle \iiint _{T}\rho ^{4}\sin \theta \,d\rho \,d\theta \,d\varphi =\int _{0}^{\pi }\sin \varphi \,d\varphi \int _{0}^{4}\rho ^{4}d\rho \int _{0}^{2\pi }d\theta =2\pi \int _{0}^{\pi }\sin \varphi \left[{\frac {\rho ^{5}}{5}}\right]_{0}^{4}\,d\varphi =2\pi \left[{\frac {\rho ^{5}}{5}}\right]_{0}^{4}{\Big [}-\cos \varphi {\Big ]}_{0}^{\pi }={\frac {4096\pi }{5}}.}

Example 4c. The domain D is the ball with center at the origin and radius 3a,

D=\left\{x^{2}+y^{2}+z^{2}\leq 9a^{2}\right\}

and f(x, y, z) = x2 + y2 is the function to integrate.

Looking at the domain, it seems convenient to adopt the passage to spherical coordinates, in fact, the intervals of the variables that delimit the new T region are obviously:

{\displaystyle T=\{0\leq \rho \leq 3a,\ 0\leq \varphi \leq 2\pi ,\ 0\leq \theta \leq \pi \}.}

However, applying the transformation, we get

{\displaystyle f(x,y,z)=x^{2}+y^{2}\longrightarrow \rho ^{2}\sin ^{2}\theta \cos ^{2}\varphi +\rho ^{2}\sin ^{2}\theta \sin ^{2}\varphi =\rho ^{2}\sin ^{2}\theta .}

Applying the formula for integration we obtain:

{\displaystyle \iiint _{T}\rho ^{2}\sin ^{2}\theta \rho ^{2}\sin \theta \,d\rho \,d\theta \,d\varphi =\iiint _{T}\rho ^{4}\sin ^{3}\theta \,d\rho \,d\theta \,d\varphi }

which can be solved by turning it into an iterated integral.


{\displaystyle \iiint _{T}\rho ^{4}\sin ^{3}\theta \,d\rho \,d\theta \,d\varphi =\underbrace {\int _{0}^{3a}\rho ^{4}d\rho } _{I}\,\underbrace {\int _{0}^{\pi }\sin ^{3}\theta \,d\theta } _{II}\,\underbrace {\int _{0}^{2\pi }d\varphi } _{III}}.

{\displaystyle I=\left.\int _{0}^{3a}\rho ^{4}d\rho ={\frac {\rho ^{5}}{5}}\right\vert _{0}^{3a}={\frac {243}{5}}a^{5}},

{\displaystyle II=\int _{0}^{\pi }\sin ^{3}\theta \,d\theta =-\int _{0}^{\pi }\sin ^{2}\theta \,d(\cos \theta )=\int _{0}^{\pi }(\cos ^{2}\theta -1)\,d(\cos \theta )=\left.{\frac {\cos ^{3}\theta }{3}}\right|_{0}^{\pi }-\left.\cos \theta \right|_{0}^{\pi }={\frac {4}{3}}},

{\displaystyle III=\int _{0}^{2\pi }d\varphi =2\pi }.


Collecting all parts,

{\displaystyle \iiint _{T}\rho ^{4}\sin ^{3}\theta \,d\rho \,d\theta \,d\varphi =I\cdot II\cdot III={\frac {243}{5}}a^{5}\cdot {\frac {4}{3}}\cdot 2\pi ={\frac {648}{5}}\pi a^{5}}.


Alternatively, this problem can be solved by using the passage to cylindrical coordinates. The new T intervals are

{\displaystyle T=\left\{0\leq \rho \leq 3a,\ 0\leq \varphi \leq 2\pi ,\ -{\sqrt {9a^{2}-\rho ^{2}}}\leq z\leq {\sqrt {9a^{2}-\rho ^{2}}}\right\};}

the z interval has been obtained by dividing the ball into two hemispheres simply by solving the inequality from the formula of D (and then directly transforming x2 + y2 into ρ2). The new function is simply ρ2. Applying the integration formula

{\displaystyle \iiint _{T}\rho ^{2}\rho \,d\rho \,d\varphi \,dz.}

Then we get

Sours: https://en.wikipedia.org/wiki/Multiple_integral
Change of Variables \u0026 The Jacobian - Multi-variable Integration
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Section 4-2 : Iterated Integrals

In the previous section we gave the definition of the double integral. However, just like with the definition of a single integral the definition is very difficult to use in practice and so we need to start looking into how we actually compute double integrals. We will continue to assume that we are integrating over the rectangle

\[R = \left[ {a,b} \right] \times \left[ {c,d} \right]\]

We will look at more general regions in the next section.

The following theorem tells us how to compute a double integral over a rectangle.

Fubini’s Theorem

If \(f\left( {x,y} \right)\) is continuous on \(R = \left[ {a,b} \right] \times \left[ {c,d} \right]\) then,

\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \int_{{\,a}}^{{\,b}}{{\int_{{\,c}}^{{\,d}}{{f\left( {x,y} \right)\,dy}}\,dx}} = \int_{{\,c}}^{{\,d}}{{\int_{{\,a}}^{{\,b}}{{f\left( {x,y} \right)\,dx}}\,dy}}\]

These integrals are called iterated integrals.

Note that there are in fact two ways of computing a double integral over a rectangle and also notice that the inner differential matches up with the limits on the inner integral and similarly for the outer differential and limits. In other words, if the inner differential is \(dy\) then the limits on the inner integral must be \(y\) limits of integration and if the outer differential is \(dy\) then the limits on the outer integral must be \(y\) limits of integration.

Now, on some level this is just notation and doesn’t really tell us how to compute the double integral. Let’s just take the first possibility above and change the notation a little.

\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \int_{{\,a}}^{{\,b}}{{\left[ {\int_{{\,c}}^{{\,d}}{{f\left( {x,y} \right)\,dy}}} \right]\,dx}}\]

We will compute the double integral by first computing

\[\int_{{\,c}}^{{\,d}}{{f\left( {x,y} \right)\,dy}}\]

and we compute this by holding \(x\) constant and integrating with respect to \(y\) as if this were a single integral. This will give a function involving only \(x\)’s which we can in turn integrate.

We’ve done a similar process with partial derivatives. To take the derivative of a function with respect to \(y\) we treated the \(x\)’s as constants and differentiated with respect to \(y\) as if it was a function of a single variable.

Double integrals work in the same manner. We think of all the \(x\)’s as constants and integrate with respect to \(y\) or we think of all \(y\)’s as constants and integrate with respect to \(x\).

Let’s take a look at some examples.

Example 1Compute each of the following double integrals over the indicated rectangles.
  1. \( \displaystyle \iint\limits_{R}{{6x{y^2}\,dA}}\), \(R = \left[ {2,4} \right] \times \left[ {1,2} \right]\)
  2. \( \displaystyle \iint\limits_{R}{{2x - 4{y^3}\,dA}}\), \(R = \left[ { - 5,4} \right] \times \left[ {0,3} \right]\)
  3. \( \displaystyle \iint\limits_{R}{{{x^2}{y^2} + \cos \left( {\pi x} \right) + \sin \left( {\pi y} \right)\,dA}}\), \(R = \left[ { - 2, - 1} \right] \times \left[ {0,1} \right]\)
  4. \( \displaystyle \iint\limits_{R}{{\frac{1}{{{{\left( {2x + 3y} \right)}^2}}}\,dA}}\), \(R = \left[ {0,1} \right] \times \left[ {1,2} \right]\)
  5. \( \displaystyle \iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}}\), \(R = \left[ { - 1,2} \right] \times \left[ {0,1} \right]\)
Show All Solutions Hide All Solutions
a\( \displaystyle \iint\limits_{R}{{6x{y^2}\,dA}}\), \(R = \left[ {2,4} \right] \times \left[ {1,2} \right]\) Show Solution

It doesn’t matter which variable we integrate with respect to first, we will get the same answer regardless of the order of integration. To prove that let’s work this one with each order to make sure that we do get the same answer.

Solution 1
In this case we will integrate with respect to \(y\) first. So, the iterated integral that we need to compute is,

\[\iint\limits_{R}{{6x{y^2}\,dA}} = \int_{{\,2}}^{{\,4}}{{\int_{{\,1}}^{{\,2}}{{6x{y^2}\,dy}}\,dx}}\]

When setting these up make sure the limits match up to the differentials. Since the \(dy\) is the inner differential (i.e. we are integrating with respect to \(y\) first) the inner integral needs to have \(y\) limits.

To compute this we will do the inner integral first and we typically keep the outer integral around as follows,

\[\begin{align*}\iint\limits_{R}{{6x{y^2}\,dA}} & = \int_{{\,2}}^{{\,4}}{{\left. {\left( {2x{y^3}} \right)} \right|_1^2\,dx}}\\ & = \int_{{\,2}}^{{\,4}}{{16x - 2x\,dx}}\\ & = \int_{{\,2}}^{{\,4}}{{14x\,dx}}\end{align*}\]

Remember that we treat the \(x\) as a constant when doing the first integral and we don’t do any integration with it yet. Now, we have a normal single integral so let’s finish the integral by computing this.

\[\iint\limits_{R}{{6x{y^2}\,dA}} = \left. {7{x^2}} \right|_2^4 = 84\]

Solution 2In this case we’ll integrate with respect to \(x\) first and then \(y\). Here is the work for this solution.

\[\begin{align*}\iint\limits_{R}{{6x{y^2}\,dA}} & = \int_{{\,1}}^{{\,2}}{{\int_{{\,2}}^{{\,4}}{{6x{y^2}\,dx}}\,dy}}\\ & = \int_{{\,1}}^{{\,2}}{{\left. {\left( {3{x^2}{y^2}} \right)} \right|_2^4\,dy}}\\ & = \int_{{\,1}}^{{\,2}}{{36{y^2}\,dy}}\\ & = \left. {12{y^3}} \right|_1^2\\ & = 84\end{align*}\]

Sure enough the same answer as the first solution.

So, remember that we can do the integration in any order.


b\( \displaystyle \iint\limits_{R}{{2x - 4{y^3}\,dA}}\), \(R = \left[ { - 5,4} \right] \times \left[ {0,3} \right]\) Show Solution

For this integral we’ll integrate with respect to \(y\) first.

\[\begin{align*}\iint\limits_{R}{{2x - 4{y^3}\,dA}} & = \int_{{\, - 5}}^{{\,4}}{{\int_{{\,0}}^{{\,3}}{{2x - 4{y^3}\,dy}}\,dx}}\\ & = \int_{{\, - 5}}^{{\,4}}{{\left. {\left( {2xy - {y^4}} \right)} \right|_0^3\,dx}}\\ & = \int_{{\, - 5}}^{{\,4}}{{6x - 81\,dx}}\\ & = \left. {\left( {3{x^2} - 81x} \right)} \right|_{ - 5}^4\\ & = - 756\end{align*}\]

Remember that when integrating with respect to \(y\) all \(x\)’s are treated as constants and so as far as the inner integral is concerned the 2\(x\) is a constant and we know that when we integrate constants with respect to \(y\) we just tack on a \(y\) and so we get \(2xy\) from the first term.


c\( \displaystyle \iint\limits_{R}{{{x^2}{y^2} + \cos \left( {\pi x} \right) + \sin \left( {\pi y} \right)\,dA}}\), \(R = \left[ { - 2, - 1} \right] \times \left[ {0,1} \right]\) Show Solution

In this case we’ll integrate with respect to \(x\) first.

\[\begin{align*}\iint\limits_{R}{{{x^2}{y^2} + \cos \left( {\pi x} \right) + \sin \left( {\pi y} \right)\,dA}} & = \int_{{\,0}}^{{\,1}}{{\int_{{\, - 2}}^{{\, - 1}}{{{x^2}{y^2} + \cos \left( {\pi x} \right) + \sin \left( {\pi y} \right)\,dx}}\,dy}}\\ & = \int_{{\,0}}^{{\,1}}{{\left. {\left( {\frac{1}{3}{x^3}{y^2} + \frac{1}{\pi }\sin \left( {\pi x} \right) + x\sin \left( {\pi y} \right)} \right)} \right|_{ - 2}^{ - 1}\,dy}}\\ & = \int_{{\,0}}^{{\,1}}{{\frac{7}{3}{y^2} + \sin \left( {\pi y} \right)\,dy}}\\ & = \left. {\frac{7}{9}{y^3} - \frac{1}{\pi }\cos \left( {\pi y} \right)} \right|_0^1\\ & = \frac{7}{9} + \frac{2}{\pi }\end{align*}\]

Don’t forget your basic Calculus I substitutions!


d\( \displaystyle \iint\limits_{R}{{\frac{1}{{{{\left( {2x + 3y} \right)}^2}}}\,dA}}\), \(R = \left[ {0,1} \right] \times \left[ {1,2} \right]\) Show Solution

In this case because the limits for \(x\) are kind of nice (i.e. they are zero and one which are often nice for evaluation) let’s integrate with respect to \(x\) first. We’ll also rewrite the integrand to help with the first integration.

\[\begin{align*}\iint\limits_{R}{{{{\left( {2x + 3y} \right)}^{ - 2}}\,dA}} & = \int_{{\,1}}^{{\,2}}{{\int_{{\,0}}^{{\,1}}{{{{\left( {2x + 3y} \right)}^{ - 2}}\,dx}}\,dy}}\\ & = \int_{{\,1}}^{{\,2}}{{\left. {\left( { - \frac{1}{2}{{\left( {2x + 3y} \right)}^{ - 1}}} \right)} \right|_0^1\,dy}}\\ & = - \frac{1}{2}\int_{{\,1}}^{{\,2}}{{\frac{1}{{2 + 3y}} - \frac{1}{{3y}}\,dy}}\\ & = - \left. {\frac{1}{2}\left( {\frac{1}{3}\ln \left| {2 + 3y} \right| - \frac{1}{3}\ln \left| y \right|} \right)} \right|_1^2\\ & = - \frac{1}{6}\left( {\ln 8 - \ln 2 - \ln 5} \right)\end{align*}\]

e\( \displaystyle \iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}}\), \(R = \left[ { - 1,2} \right] \times \left[ {0,1} \right]\) Show Solution

Now, while we can technically integrate with respect to either variable first sometimes one way is significantly easier than the other way. In this case it will be significantly easier to integrate with respect to \(y\) first as we will see.

\[\iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}} = \int_{{ - 1}}^{2}{{\int_{0}^{1}{{x{{\bf{e}}^{xy}}\,dy}}\,dx}}\]

The \(y\) integration can be done with the quick substitution,

\[u = xy\hspace{0.5in}du = x\,dy\]

which gives

\[\begin{align*}\iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}} & = \int_{{ - 1}}^{2}{{\left. {{{\bf{e}}^{xy}}} \right|_0^1\,dx}}\\ & = \int_{{ - 1}}^{2}{{{{\bf{e}}^x} - 1\,dx}}\\ & = \left. {\left( {{{\bf{e}}^x} - x} \right)} \right|_{ - 1}^2\\ & = {{\bf{e}}^2} - 2 - \left( {{{\bf{e}}^{ - 1}} + 1} \right)\\ & = {{\bf{e}}^2} - {{\bf{e}}^{ - 1}} - 3\end{align*}\]

So, not too bad of an integral there provided you get the substitution. Now let’s see what would happen if we had integrated with respect to \(x\) first.

\[\iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}} = \int_{0}^{1}{{\int_{{ - 1}}^{2}{{x{{\bf{e}}^{xy}}\,dx}}\,dy}}\]

In order to do this we would have to use integration by parts as follows,

\[\begin{align*}u & = x & \hspace{0.25in} dv & = {{\bf{e}}^{xy}}\,dx\\ du & = dx & \hspace{0.25in} v &= \frac{1}{y}{{\bf{e}}^{xy}}\end{align*}\]

The integral is then,

\[\begin{align*}\iint\limits_{R}{{x{{\bf{e}}^{xy}}\,dA}} & = \int_{0}^{1}{{\left. {\left( {\frac{x}{y}{{\bf{e}}^{xy}} - \int_{{}}^{{}}{{\frac{1}{y}{{\bf{e}}^{xy}}\,dx}}} \right)} \right|_{ - 1}^2\,dy}}\\ & = \int_{0}^{1}{{\left. {\left( {\frac{x}{y}{{\bf{e}}^{xy}} - \frac{1}{{{y^2}}}{{\bf{e}}^{xy}}} \right)} \right|_{ - 1}^2\,dy}}\\ & = \int_{0}^{1}{{\left( {\frac{2}{y}{{\bf{e}}^{2y}} - \frac{1}{{{y^2}}}{{\bf{e}}^{2y}}} \right) - \left( { - \frac{1}{y}{{\bf{e}}^{ - y}} - \frac{1}{{{y^2}}}{{\bf{e}}^{ - y}}} \right)\,dy}}\end{align*}\]

We’re not even going to continue here as these are very difficult, if not impossible, integrals to do.

As we saw in the previous set of examples we can do the integral in either direction. However, sometimes one direction of integration is significantly easier than the other so make sure that you think about which one you should do first before actually doing the integral.

The next topic of this section is a quick fact that can be used to make some iterated integrals somewhat easier to compute on occasion.

There is a nice special case of this kind of integral. First, let’s assume that \(f\left( {x,y} \right) = g\left( x \right)h\left( y \right)\) and let’s also assume we are integrating over a rectangle given by \(R = \left[ {a,b} \right] \times \left[ {c,d} \right]\). Then, the integral becomes,

\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \iint\limits_{R}{{g\left( x \right)h\left( y \right)\,dA}} = \int_{c}^{d}{{\int_{a}^{b}{{g\left( x \right)h\left( y \right)\,dx}}\,dy}}\]

Note that it doesn’t matter in this case which variable we integrate first as either order will arrive at the same result with the same work.

Next, notice that because the inner integral is with respect to \(x\) and \(h\left( y \right)\) is a function only of \(y\) it can be considered a “constant” as far as the \(x\) integration is concerned (changing \(x\) will not affect the value of \(y\)!) and because it is also times \(g\left( x \right)\) we can factor the \(h\left( y \right)\) out of the inner integral. Doing this gives,

\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \iint\limits_{R}{{g\left( x \right)h\left( y \right)\,dA}} = \int_{c}^{d}{{h\left( y \right)\int_{a}^{b}{{g\left( x \right)\,dx}}\,dy}}\]

Now, \(\int_{a}^{b}{{g\left( x \right)\,dx}}\) is a standard Calculus I definite integral and we know that its value is just a constant. Therefore, it can be factored out of the \(y\) integration to get,

\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \iint\limits_{R}{{g\left( x \right)h\left( y \right)\,dA}} = \int_{a}^{b}{{g\left( x \right)\,dx}}\int_{c}^{d}{{h\left( y \right)\,dy}}\]

In other words, if we can break up the function into a function only of \(x\) times a function of only \(y\) then we can do the two integrals individually and multiply them together.

Here is a quick summary of this idea.

Fact

If \(f\left( {x,y} \right) = g\left( x \right)h\left( y \right)\) and we are integrating over the rectangle \(R = \left[ {a,b} \right] \times \left[ {c,d} \right]\) then,

\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} = \iint\limits_{R}{{g\left( x \right)h\left( y \right)\,dA}} = \left( {\int_{{\,a}}^{{\,b}}{{g\left( x \right)\,dx}}} \right)\left( {\int_{{\,c}}^{{\,d}}{{h\left( y \right)\,dy}}} \right)\]

Let’s do a quick example using this integral.

Example 2Evaluate \( \displaystyle \iint\limits_{R}{{x{{\cos }^2}\left( y \right)\,dA}}\), \(R = \left[ { - 2,3} \right] \times \left[ {0,\frac{\pi }{2}} \right]\).
Show Solution

Since the integrand is a function of \(x\) times a function of \(y\) we can use the fact.

\[\begin{align*}\iint\limits_{R}{{x{{\cos }^2}\left( y \right)\,dA}} & = \left( {\int_{{\, - 2}}^{{\,3}}{{x\,dx}}} \right)\left( {\int_{{\,0}}^{{\,\frac{\pi }{2}}}{{{{\cos }^2}\left( y \right)\,dy}}} \right)\\ & = \left. {\left( {\frac{1}{2}{x^2}} \right)} \right|_{ - 2}^3\left( {\frac{1}{2}\int_{{\,0}}^{{\,\frac{\pi }{2}}}{{1 + \cos \left( {2y} \right)\,dy}}} \right)\\ & = \left( {\frac{5}{2}} \right)\left( {\frac{1}{2}\left. {\left( {y + \frac{1}{2}\sin \left( {2y} \right)} \right)} \right|_0^{\frac{\pi }{2}}} \right)\\ & = \frac{{5\pi }}{8}\end{align*}\]

We have one more topic to discuss in this section. This topic really doesn’t have anything to do with iterated integrals, but this is as good a place as any to put it and there are liable to be some questions about it at this point as well so this is as good a place as any.

What we want to do is discuss single indefinite integrals of a function of two variables. In other words, we want to look at integrals like the following.

\[\begin{align*}& \int{{x{{\sec }^2}\left( {2y} \right) + 4xy\,dy}}\\ & \int{{{x^3} - {{\bf{e}}^{ - \frac{x}{y}}}\,dx}}\end{align*}\]

From Calculus I we know that these integrals are asking what function that we differentiated to get the integrand. However, in this case we need to pay attention to the differential (\(dy\) or \(dx\)) in the integral, because that will change things a little.

In the case of the first integral we are asking what function we differentiated with respect to \(y\) to get the integrand while in the second integral we’re asking what function differentiated with respect to \(x\) to get the integrand. For the most part answering these questions isn’t that difficult. The important issue is how we deal with the constant of integration.

Here are the integrals.

\[\begin{align*}\int{{x{{\sec }^2}\left( {2y} \right) + 4xy\,dy}} & = \frac{x}{2}\tan \left( {2y} \right) + 2x{y^2} + g\left( x \right)\\ \int{{{x^3} - {{\bf{e}}^{ - \frac{x}{y}}}\,dx}} & = \frac{1}{4}{x^4} + y\,{{\bf{e}}^{ - \frac{x}{y}}} + h\left( y \right)\end{align*}\]

Notice that the “constants” of integration are now functions of the opposite variable. In the first integral we are differentiating with respect to \(y\) and we know that any function involving only \(x\)’s will differentiate to zero and so when integrating with respect to \(y\) we need to acknowledge that there may have been a function of only \(x\)’s in the function and so the “constant” of integration is a function of \(x\).

Likewise, in the second integral, the “constant” of integration must be a function of \(y\) since we are integrating with respect to \(x\). Again, remember if we differentiate the answer with respect to \(x\) then any function of only \(y\)’s will differentiate to zero.

Sours: https://tutorial.math.lamar.edu/classes/calciii/IteratedIntegrals.aspx

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