Change of Variables

Learning Objectives

Evaluate a double integral using a change of variables.
Evaluate a triple integral using a change of variables.

Change of Variables for Double Integrals

We have already seen that, under the change of variables $T(u, v)=(x, y)$ where $x=g(u, v)$ and $y=h(u, v)$ , a small region ${\Delta}{A}$ in the $xy$ -plane is related to the area formed by the product ${\Delta}{u}{\Delta}{v}$ in the $uv$ -plane by the approximation

$\large{{\Delta}{A}{\approx}{J}{(u,v)}{\Delta}{u},{\Delta}{v}}$ .

Now let’s go back to the definition of double integral for a minute:

$\large{\underset{R}{\displaystyle\iint}f(x,y) \ dA=\displaystyle\lim_{m,n\to\infty}\displaystyle\sum_{i=1}^m\displaystyle\sum_{j=1}^n{f}(x_{ij},y_{ij})\Delta{A}}$ .

Referring to Figure 1, observe that we divided the region $S$ in the $uv$ -plane into small subrectangles $S_{ij}$ and we let the subrectangles $R_{ij}$ in the $xy$ -plane be the images of $S_{ij}$ under the transformation $T(u, v)=(x, y)$ .

On the left-hand side of this figure, there is a rectangle S with an inscribed red oval and a subrectangle with lower right corner point (u sub ij, v sub ij), height Delta v, and length Delta u given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with T. On the right-hand side of this figure there is a region R with inscribed (deformed) red oval and a subrectangle R sub ij with corner point (x sub ij, y sub ij) given in the Cartesian x y-plane. The subrectangle is blown up and shown with vectors pointing along the edge from the corner point.

Figure 1. The subrectangles $S_{ij}$ in the $uv$ -plane transform into subrectangles $R_{ij}$ in the $xy$ -plane.

Then the double integral becomes

$\large{\underset{R}{\displaystyle\iint}f(x,y) \ dA=\displaystyle\lim_{m,n\to\infty}\displaystyle\sum_{i=1}^m\displaystyle\sum_{j=1}^n{f}(x_{ij},y_{ij})\Delta{A}=\displaystyle\lim_{m,n\to\infty}\displaystyle\sum_{i=1}^m\displaystyle\sum_{j=1}^nf(g(u_{ij},v_{ij}),h(u_{ij},v_{ij}))|J(u_{ij},v_{ij})|\Delta{u}\Delta{v}}$ .

Notice this is exactly the double Riemann sum for the integral

$\large{\underset{S}{\displaystyle\iint}f(g(u,v),h(u,v))\Bigg|\frac{\partial(x,y)}{\partial(u,v)}\Bigg|du \ dv}$ .

theorem: change of variables for double integrals

Let $T(u, v)=(x, y)$ where $x=g(u, v)$ and $y=h(u, v)$ be a one-to-one $C1$ transformation, with a nonzero Jacobian on the interior of the region $S$ in the $uv$ -plane; it maps $S$ into the region $R$ in the $xy$ -plane. If $f$ is continuous on $R$ , then

$\large{\underset{R}{\displaystyle\iint}f(x,y) \ dA=\underset{S}{\displaystyle\iint}f(g(u,v),h(u,v))\Bigg|\frac{\partial(x,y)}{\partial(u,v)}\Bigg|du \ dv}$ .

With this theorem for double integrals, we can change the variables from $(x, y)$ to $(u, v)$ in a double integral simply by replacing

$\large{dA=dx \ dy=\Bigg|\frac{\partial(x,y)}{\partial(u,v)}\Bigg|du \ dv}$

when we use the substitutions $x=g(u, v)$ and $y=h(u, v)$ and then change the limits of integration accordingly. This change of variables often makes any computations much simpler.

Example: changing variables from rectangular to polar coordinates

Consider the integral

$\displaystyle\int_0^2\displaystyle\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2}dy \ dx$ .

Use the change of variables $x=r\cos\theta$ and $y=r\sin\theta$ , and find the resulting integral.

Show Solution

First we need to find the region of integration. This region is bounded below by $y=0$ and above by $y=\sqrt{2x-x^2}$ (see the following figure).

A semicircle in the first quadrant of the xy plane with radius 1 and center (1, 0). The equation for this curve is given as y = the square root of (2x minus x squared)

Figure 2. Changing a region from rectangular to polar coordinates.

Squaring and collecting terms, we find that the region is the upper half of the circle $x^{2}+y^{2}-2x=0$ that is, $y^{2}+(x-1)^{2}$ . In polar coordinates, the circle is $r=2\cos\theta$ so the region of integration in polar coordinates is bounded by $0\leq{r}\leq\cos\theta$ and $0\leq\theta\leq\frac{\pi}2$ .

The Jacobian is $J(r,\theta)=r$ , as shown in Example “Finding the Jacobian”. Since $r\geq0$ , we have $|J(r,\theta)|=r$ .

The integrand $\sqrt{x^2+y^2}$ changes to $r$ in polar coordinates, so the double iterated integral is

$\displaystyle\int_0^2\displaystyle\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2}dy \ dx=\displaystyle\int_0^{\pi/2}\displaystyle\int_0^{2\cos\theta}r|J(r,\theta)|dr \ d\theta=\displaystyle\int_0^{\pi/2}\displaystyle\int_0^{2\cos\theta}r^2 \ dr \ d\theta$ .

try it

Considering the integral $\displaystyle\int_0^1\displaystyle\int_0^{\sqrt{1-x^2}}(x^2+y^2)dy \ dx$ , use the change of variables $x=r\cos\theta$ and $y=\sin\theta$ , and find the resulting integral.

Show Solution

$\displaystyle\int_0^{\pi/2}\displaystyle\int_0^1r^3 \ dr \ d\theta$

Notice in the next example that the region over which we are to integrate may suggest a suitable transformation for the integration. This is a common and important situation.

Example: changing variables

Consider the integral $\underset{R}{\displaystyle\iint}(x-y)dy \ dx$ , where $R$ is the parallelogram joining the points $(1, 2)$ , $(3, 4)$ , $(4, 3)$ , and $(6, 5)$ (Figure 3). Make appropriate changes of variables, and write the resulting integral.

Figure 3. The region of integration for the given integral.

Show Solution

First, we need to understand the region over which we are to integrate. The sides of the parallelogram are $x-y+1=0$ , $x-y-1=0$ , $x-3y+5=0$ , and $x-3y+9=0$ (Figure 4). Another way to look at them is $x-y=-1$ , $x-y=1$ , $x-3y=-5$ , and $x-3y=-9$ .

Clearly the parallelogram is bounded by the lines $y=x+1$ , $y=x-1$ , $y=\frac13(x+5)$ , and $y=\frac13(x+9)$ .

Notice that if we were to make $u=x-y$ and $v=x-3y$ , then the limits on the integral would be $-1\leq{u}\leq1$ and $-9\leq{v}\leq-5$ .

To solve for $x$ and $y$ , we multiply the first equation by $3$ and subtract the second equation, $3u-v=(3x-3y)-(x-3y)=2x$ . Then we have $x=\frac{3u-v}2$ . Moreover, if we simply subtract the second equation from the first, we get $u-v=(x-y)-(x-3y)=2y$ and $y=\frac{u-v}2$ .

A parallelogram R with corners (1, 2), (3, 4), (6, 5), and (4, 3) formed by the lines y = x + 1, y = x minus 1, y = (x + 9)/3, and y = (x + 5)/3.

Figure 4. A parallelogram in the $xy$ -plane that we want to transform by a change in variables.

Thus, we can choose the transformation

$\large{T(u,v)=\left(\frac{3u-v}2,\frac{u-v}2\right)}$

and compute the Jacobian $J(u, v)$ . We have

$\large{J(u,v)=\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}\frac{\partial{x}}{\partial{u}}&\frac{\partial{x}}{\partial{v}} \\ \frac{\partial{y}}{\partial{u}}&\frac{\partial{y}}{\partial{v}}\end{vmatrix}=\begin{vmatrix}3/2&-1/2 \\ 1/2&-1/2\end{vmatrix}=-\frac34+\frac14=-\frac12}$ .

Therefore, $|J(u,v)|=\frac12$ . Also, the original integrand becomes

$\large{x-y=\frac12[3u-v-u+v]=\frac12[3u-u]=\frac12[2u]=u}$ .

Therefore, by the use of the transformation $T$ , the integral changes to

$\large{\underset{R}{\displaystyle\iint}(x-y)dy \ dx=\displaystyle\int_{-9}^{-5}\displaystyle\int_{-1}^{1}J(u,v)u \ du \ dv=\displaystyle\int_{-9}^{-5}\displaystyle\int_{-1}^{1}\left(\frac12\right)u \ du \ dv}$ ,

which is much simpler to compute. In fact, it is easily found to be zero. And this is just one example of why we transform integrals like this.

try it

Make appropriate changes of variables in the integral $\underset{R}{\displaystyle\iint}\frac4{(x-y)^2}dy \ dx$ , where $R$ is the trapezoid bounded by the lines $x-y=2$ , $x-y=4$ , $x=0$ , and $y=0$ . Write the resulting integral.

Show Solution

$x=\frac12(v+u)$ and $y=\frac12(u-v)$ and $\displaystyle\int_2^4\displaystyle\int_{-u}^u\frac4{u^2}\left(\frac12\right)dv \ du$ .

We are ready to give a problem-solving strategy for change of variables.

Problem solving strategy: change of variables

Sketch the region given by the problem in the $xy$ -plane and then write the equations of the curves that form the boundary.
Depending on the region or the integrand, choose the transformations $x=g(u, v)$ and $y=h(u, v)$ .
Determine the new limits of integration in the $uv$ -plane.
Find the Jacobian $J(u, v)$ .
In the integrand, replace the variables to obtain the new integrand.
Replace $dy \ dx$ or $dx \ dy$ , whichever occurs, by $J(u,v)du \ dv$ .

In the next example, we find a substitution that makes the integrand much simpler to compute.

Example: evaluating an integral

Using the change of variables $u=x-y$ and $v=x+y$ , evaluate the integral

$\large{\underset{R}{\displaystyle\iint}(x-y)e^{x^2-y^2}dA}$ ,

where $R$ is the region bounded by the lines $x+y=1$ and $x+y=3$ and the curves $x^{2}-y^{2}=-1$ and $x^{2}-y^{2}=1$ (see the first region in Figure 5).

Show Solution

As before, first find the region $R$ and picture the transformation so it becomes easier to obtain the limits of integration after the transformations are made (Figure 5).

Figure 5. Transforming the region $R$ into the region $S$ to simplify the computation of an integral.

Given $u=x-y$ and $v=x+y$ , we have $x=\frac{u+v}2$ and $y=\frac{v-u}2$ and hence the transformation to use is $T(u,v)=\left(\frac{u+v}2,\frac{v-u}2\right)$ . The lines $x+y=1$ and $x+y=3$ become $v=1$ and $v=3$ , respectively. The curves $x^{2}-y^{2}=1$ and $x^{2}-y^{2}=-1$ become $uv=1$ and $uv=-1$ , respectively.

Thus we can describe the region $S$ (see the second region Figure 5) as

$\large{s=\left\{(u,v)|1\leq{v}\leq3,\frac{-1}v\leq{u}\leq\frac1v\right\}}$ .

The Jacobian for this transformation is

Therefore, by using the transformation $T$ , the integral changes to

$\large{\underset{R}{\displaystyle\iint}(x-y)e^{x^2-y^2}dA=\frac12\displaystyle\int_1^3\displaystyle\int_{-1/v}^{1/v}ue^{uv}du \ dv}$ .

Doing the evaluation, we have

$\large{\frac12\displaystyle\int_1^3\displaystyle\int_{-1/v}^{1/v}ue^{uv}du \ dv=\frac4{3e}\approx0.490}$ .

try it

Using the substitutions $x=v$ and $y=\sqrt{u+v}$ , evaluate the integral $\underset{R}{\displaystyle\iint}y\sin(y^2-x)dA$ where $R$ is the region bounded by the lines $y=\sqrt{x},\ x=2$ and $y=0$ .

Show Solution

$\frac12\sin{2}-1$

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.47” here (opens in new window).

Change of Variables for Triple Integrals

Changing variables in triple integrals works in exactly the same way. Cylindrical and spherical coordinate substitutions are special cases of this method, which we demonstrate here.

Suppose that $G$ is a region in $uvw$ -space and is mapped to $D$ in $xyz$ -space (Figure 6) by a one-to-one $C1$ transformation $T(u, v, w)=(x, y, z)$ where $x=g(u, v, w)$ , $y=h(u, v, w)$ , and $z=k(u, v, w)$ .

On the left-hand side of this figure, there is a region G in u v w space. Then there is an arrow from this graph to the right-hand side of the figure marked with x = g(u, v, w), y = h(u, v, w), and z = k(u, v, w). On the right-hand side of this figure there is a region D in xyz space.

Figure 6. A region $G$ in $uvw$ -space mapped to a region $D$ in $xyz$ -space.

Then any function $F(x, y, z)$ defined on $D$ can be thought of as another function $H(u, v, w)$ that is defined on $G$ :

$\large{F(x,y,z)=F(g(u,v,w), \ h(u,v,w) \ k(u,v,w))= \ H(u,v,w)}$ .

Now we need to define the Jacobian for three variables.

definition

The Jacobian determinant $J(u, v, w)$ in three variables is defined as follows:

$\large{J(u,v,w)=\begin{vmatrix}\frac{\partial{x}}{\partial{u}}&\frac{\partial{y}}{\partial{u}}&\frac{\partial{z}}{\partial{u}} \\ \frac{\partial{x}}{\partial{v}}&\frac{\partial{y}}{\partial{v}}&\frac{\partial{z}}{\partial{v}} \\ \frac{\partial{x}}{\partial{w}}&\frac{\partial{y}}{\partial{w}}&\frac{\partial{z}}{\partial{w}}\end{vmatrix}}$ .

This is also the same as

$\large{J(u,v,w)=\begin{vmatrix}\frac{\partial{x}}{\partial{u}}&\frac{\partial{x}}{\partial{v}}&\frac{\partial{x}}{\partial{w}} \\ \frac{\partial{y}}{\partial{u}}&\frac{\partial{y}}{\partial{v}}&\frac{\partial{y}}{\partial{w}} \\ \frac{\partial{z}}{\partial{u}}&\frac{\partial{z}}{\partial{v}}&\frac{\partial{z}}{\partial{w}}\end{vmatrix}}$ .

The Jacobian can also be simply denoted as $\frac{\partial(x,y,z)}{\partial(u,v,w)}$ .

With the transformations and the Jacobian for three variables, we are ready to establish the theorem that describes change of variables for triple integrals.

theorem: change of variables for triple integrals

Let $T(u, v, w)=(x, y, z)$ where $x=g(u, v, w)$ , $y=h(u, v, w)$ , and $z=k(u, v, w)$ , be a one-to-one $C1$ transformation, with a nonzero Jacobian, that maps the region $G$ in the $uvw$ -plane into the region $D$ in the $xyz$ -plane. As in the two-dimensional case, if $F$ is continuous on $D$ , then

$\hspace{3cm}\large{\begin{align} \underset{R}{\displaystyle\iiint}F(x,y,z)dV&=\underset{G}{\displaystyle\iiint}F(g(u,v,w), \ h(u,v,w), \ k(u,v,w))\Bigg|\frac{\partial(x,y,z)}{\partial(u,v,w)}\Bigg|du \ dv \ dw \\ &=\underset{G}{\displaystyle\iiint}H(u,v,w)|J(u,v,w)|du \ dv \ dw \end{align}}$ .

Let us now see how changes in triple integrals for cylindrical and spherical coordinates are affected by this theorem. We expect to obtain the same formulas as in Triple Integrals in Cylindrical and Spherical Coordinates.

Example: obtaining formulas in triple integrals for cylindrical and spherical coordinates

Derive the formula in triple integrals for

cylindrical and
spherical coordinates.

Show Solution

For cylindrical coordinates, the transformation is $T(r,\theta,z)=(x,y,z)$ from the Cartesian $r\theta{z}$ –plane to the Cartesian $xyz$ -plane (Figure 7). Here $x=r\cos\theta$ , $y=r\sin\theta$ , and $z=z$ . The Jacobian for the transformation is
$\hspace{1.5cm}\large{\begin{align}J(r,\theta,z)&=\frac{\partial(x,y,z)}{\partial(r,\theta,z)}=\begin{vmatrix}\frac{\partial{x}}{\partial{r}}&\frac{\partial{x}}{\partial{\theta}}&\frac{\partial{x}}{\partial{z}} \\ \frac{\partial{y}}{\partial{r}}&\frac{\partial{y}}{\partial{\theta}}&\frac{\partial{y}}{\partial{z}} \\ \frac{\partial{z}}{\partial{r}}&\frac{\partial{z}}{\partial{\theta}}&\frac{\partial{z}}{\partial{z}}\end{vmatrix} \\&=\begin{vmatrix}\cos\theta&-r\sin\theta&0 \\ \sin\theta&r\cos\theta&0 \\ 0&0&1\end{vmatrix}=r\cos^2\theta+r\sin^2\theta=r(\cos^2\theta+\sin^s\theta)=r \end{align}}$ .
We know that $r\geq0$ , so $|J(r,\theta,z)|=r$ . Then the triple integral is

$\large{\underset{D}{\displaystyle\iiint}f(x,y,z)dV=\underset{G}{\displaystyle\iiint}f(r\cos\theta,r\sin\theta,z)r \ dr \ d\theta \ dz}$ .

Figure 7. The transformation from rectangular coordinates to cylindrical coordinates can be treated as a change of variables from region $G$ in $r\theta{z}$ -space to region $D$ in $xyz$ -space.
For spherical coordinates, the transformation is $T(\rho,\theta,\varphi)=(x,y,z)$ from the Cartesian $p\theta\varphi$ –plane to the Cartesian $xyz$ –plane (Figure 8). Here $x=\rho\sin\varphi\cos\theta$ , $y=\rho\sin\varphi\sin\theta$ , and $z=\rho\cos\varphi$ . The Jacobian for the transformation is

$\large{J(\rho,\theta,\varphi)=\frac{\partial{(x,y,z)}}{\partial{(\rho,\theta\varphi)}}=\begin{vmatrix}\frac{\partial{x}}{\partial{\rho}}&\frac{\partial{x}}{\partial{\theta}}&\frac{\partial{x}}{\partial{\varphi}} \\ \frac{\partial{y}}{\partial{\rho}}&\frac{\partial{y}}{\partial{\theta}}&\frac{\partial{y}}{\partial{\varphi}} \\ \frac{\partial{z}}{\partial{\rho}}&\frac{\partial{z}}{\partial{\theta}}&\frac{\partial{z}}{\partial{\varphi}}\end{vmatrix}=\begin{vmatrix}\sin\varphi\cos\theta&-\rho\sin\varphi\sin\theta&\rho\cos\varphi\cos\theta \\ \sin\varphi\sin\theta&-\rho\sin\varphi\cos\theta&\rho\cos\varphi\sin\theta \\ \cos\theta&0&-\rho\sin\varphi\end{vmatrix}}$

Expanding the determinant with respect to the third row:
$\hspace{2cm}\begin{align} &=\cos\varphi\begin{vmatrix}-\rho\sin\varphi\sin\theta&\rho\cos\varphi\cos\theta \\ \rho\sin\varphi\sin\theta&\rho\cos\varphi\sin\theta\end{vmatrix}-\rho\sin\varphi\begin{vmatrix}\sin\varphi\cos\theta&-\rho\sin\varphi\sin\theta \\ \sin\varphi\sin\theta&\rho\sin\varphi\cos\theta\end{vmatrix} \\ &=\cos\varphi(-\rho^2\sin\varphi\cos\varphi\sin^2\theta-\rho^2\sin\varphi\cos\varphi\cos^2\theta)-\rho\sin\varphi(\rho\sin^2\varphi\cos^2\theta+\rho\sin^2\varphi\sin^2\theta) \\ &=-\rho^2\sin\varphi\cos^2\varphi(\sin^2\theta+\cos^2\theta)-\rho^2\sin\varphi\sin^2\varphi(\sin^2\theta+\cos^2\theta) \\ &=-\rho^2\sin\varphi\cos^2\varphi-\rho^2\sin\varphi\sin^2\varphi \\ &=-\rho^2\sin\varphi(\cos^2\varphi+\sin^2\varphi)=-\rho^2\sin\varphi\end{align}$
Since $0\leq\varphi\leq\pi$ , we must have $\sin\varphi\geq0$ . Thus $|J(\rho,\theta,\varphi)|=|-\rho^2\sin\varphi|=\rho^2\sin\varphi$ .

Figure 8. The transformation from rectangular coordinates to cylindrical coordinates can be treated as a change of variables from region $G$ in $\rho\theta\varphi$ -space to region $D$ in $xyz$ -space.

Then the triple integral becomes
$\underset{D}{\displaystyle\iiint}f(x,y,z)dV=\underset{G}{\displaystyle\iiint}f(\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)\rho^2\sin\varphi \ d\rho \ d\varphi \ d\theta$ .

Let’s try another example with a different substitution.

Example: evaluating a triple integral with a change of variables

Evaluate the triple integral

$\large{\displaystyle\int_0^3\displaystyle\int_0^4\displaystyle\int_{y/2}^{(y/2)+1}\left(x+\frac{z}3\right)dx \ dy \ dz}$

in $xyz$ -space by using the transformation

$\large{u=(2x-y)/2, \ v=y/2,\text{ and }w=z/3}$ .

Then integrate over an appropriate region in $uvw$ -space

Show Solution

As before, some kind of sketch of the region $G$ in $xyz$ -space over which we have to perform the integration can help identify the region $D$ in $uvw$ -space (Figure 9). Clearly $G$ in $xyz$ -space is bounded by the planes $x=y/2$ , $x=(y/2)+1$ , $y=0$ , $y=4$ , $z=0$ , and $z=4$ . We also know that we have to use $u=(2x-y)/2$ , $v=y/2$ and $w=z/3$ for the transformations. We need to solve for $x$ , $y$ , and $z$ . Here we find that $x=u+v$ , $y=2v$ , and $z=3w$ .

Using elementary algebra, we can find the corresponding surfaces for the region $G$ and the limits of integration in $uvw$ -space. It is convenient to list these equations in a table.

Equations in $xyz$ for the region $D$	Corresponding equations in $uvw$ for the region $G$	Limits for the integration in $uvw$
$x=y/2$	$u+v=2v/2=v$	$u=0$
$x=y/2$	$u+v=(2v/2)+1=v+1$	$u=1$
$y=0$	$2v=0$	$v=0$
$y=4$	$2v=4$	$v=2$
$z=0$	$3w=0$	$w=0$
$z=3$	$3w=3$	$w=1$

<img src="/apps/archive/20210823.155019/resources/9ac91afe22764bb4840cdcc0310bae65f7605c69" data-media-type="image/jpeg" alt="On the left-hand side of this figure, there is a box G with sides 1, 2, and 1 along the u, v, and w axes, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u + v, y = 2v, and z = 3w. On the right-hand side of this figure there is a region D in xyz space that is a rotated box with sides 1, 4, and 3 along the x, y, and z axes. The rear plane is marked x = y/2 or y = 2x. The front plane is marked x = y/2 + 1 or y = 2x minus 2." id="34">

Figure 9. The region $G$ in $uvw$ -space is transformed to region $D$ in $xyz$ -space

Now we can calculate the Jacobian for the transformation:

The function to be integrated becomes

$\large{f(x,y,z)=x+\frac{z}3=u+v+\frac{3w}3=u+v+w}$ .

We are now ready to put everything together and complete the problem.

$\hspace{3cm}\large{\begin{align} &\hspace{1cm}\displaystyle\int_0^3\displaystyle\int_0^4\displaystyle\int_{y/2}^{(y/2)+1}\left(x+\frac{z}3\right)dx \ dy \ dz \\ &=\displaystyle\int_0^1\displaystyle\int_0^2\displaystyle\int_0^1(u+v+w)|J(u,v,w)|du \ dv \ dw=\displaystyle\int_0^1\displaystyle\int_0^2\displaystyle\int_0^1(u+v+w)|6|du \ dv \ dw \\ &=6\displaystyle\int_0^1\displaystyle\int_0^2\displaystyle\int_0^1(u+v+w)du \ dv \ dw=6\displaystyle\int_0^1\displaystyle\int_0^2\left[\frac{u^2}2+vu+wu\right]_0^1dv \ dw \\ &=6\displaystyle\int_0^1\displaystyle\int_0^2\left(\frac12+v+w\right)dv \ dw=6\displaystyle\int_0^1\left[\frac12v+\frac{v^2}2+wv\right]_0^2dw \\ &=6\displaystyle\int_0^1(3+2w)dw=6\left[3w+w^2\right]_0^1=24 \end{align}}$ .

try it

Let $D$ be the region in $xyz$ -space defined by $1\leq{x}\leq2, \ 0\leq{xy}\leq2, \text{and} \ 0\leq{z}\leq1$ .

Evaluate $\underset{D}{\displaystyle\iiint}(x^2y+3xyz)dx \ dy \ dz$ by using the transformation $u=x$ , $v=xy$ , and $w=3z$ .

Show Solution

$\displaystyle\int_0^3\displaystyle\int_0^2\displaystyle\int_1^2\left(\frac{v}3+\frac{vw}{3u}\right)dy \ dv \ dw=2+\ln8 = 2+ 3\ln2$ .

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.48” here (opens in new window).

Module 5: Multiple Integration

Learning Objectives

Change of Variables for Double Integrals

theorem: change of variables for double integrals

Example: changing variables from rectangular to polar coordinates

try it

Example: changing variables

try it

Problem solving strategy: change of variables

Example: evaluating an integral

try it

Change of Variables for Triple Integrals

definition

theorem: change of variables for triple integrals

Example: obtaining formulas in triple integrals for cylindrical and spherical coordinates

Example: evaluating a triple integral with a change of variables

try it

Candela Citations