Multivariable calculus and vector analysis

Double integral change of variables examples

Example 1

Compute the double integral

$$\begin{array}{lll}\hfill {\iint }_{D}g\left(x,y\right)dA& & \hfill \end{array}$$

where $g\left(x,y\right)=x^2+y^2$ and $D$ is disk of radius 7 centered at origin.

Solution: Since computing this integral in rectangular coordinates is too difficult, we change to polar coordinates. (We chose polar coordinates since the disk is easily described in polar coordinates.) Let

$$\begin{array}{lll}\hfill \mathbf{T}\left(r,\theta \right)=\left(rcos\theta ,rsin\theta \right).& & \hfill \end{array}$$

be the change of variables function from polar coordinates into rectangular coordinates.

Then, by the change of variables formula, the integral becomes

$$\begin{array}{lll}\hfill {\iint }_{D}g\left(x,y\right)dA={\iint }_{D^{\ast }}g\left(\mathbf{T}\left(r,\theta \right)\right)\left|detD\mathbf{T}\left(r,\theta \right)\right|drd\theta ,& & \hfill \end{array}$$

where $D^{\ast }$ is the region in the $r\theta$-plane that is mapped by $\mathbf{T}$ onto the disk $D$. In polar coordinates, the disk is described by $0\le r\le 7$ and $0\le \theta \le 2\pi$, so this is the region $D^{\ast }$.

We calculate

$$\begin{array}{llll}\hfill g\left(\mathbf{T}\left(r,\theta \right)\right)& =g\left(rcos\theta ,rsin\theta \right)=r^2{cos}^2\theta +r^2{sin}^2\theta =r^2& \hfill & \\ \hfill \frac{\partial T_1}{\partial r}\left(r,\theta \right)& =cos\theta & \hfill & \\ \hfill \frac{\partial T_2}{\partial r}\left(r,\theta \right)& =sin\theta & \hfill & \\ \hfill \frac{\partial T_1}{\partial \theta }\left(r,\theta \right)& =-rsin\theta & \hfill & \\ \hfill \frac{\partial T_2}{\partial \theta }\left(r,\theta \right)& =rcos\theta & \hfill & \\ \hfill D\mathbf{T}\left(r,\theta \right)& =\left|\begin{array}{cc}\hfill cos\theta \hfill & \hfill sin\theta \hfill \\ \hfill -rsin\theta \hfill & \hfill rcos\theta \hfill \end{array}\right|& \hfill & \\ \hfill detD\mathbf{T}\left(r,\theta \right)& =r{cos}^2\theta -\left(-r\right){sin}^2\theta =r\left({cos}^2\theta +{sin}^2\theta \right)=r& \hfill & \end{array}$$

Since $r\ge 0$ in $D^{\ast }$, we know that $\left|detD\mathbf{T}\left(r,\theta \right)\right|=\left|r\right|=r$. Therefore the integral is

$$\begin{array}{llll}\hfill {\iint }_{D}g\left(x,y\right)dA& ={\int \; <!--nolimits\; -->}_0^{2\pi }{\int \; <!--nolimits\; -->}_0^7{r}^{2}rdrd\theta & \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_2^{2\pi }{\left.\frac{r^4}{4}\right|}_0^{7}d\theta & \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_2^{2\pi }\frac{7^4}{4}d\theta =\frac{\pi 7^4}{2}& \hfill & \end{array}$$

Note: In case you have trouble believing the formula that $dA$ becomes $\left|detD\mathbf{T}\left(r,\theta \right)\right|dr,d\theta$, we could motivate the result that, for polar coordinates, $dA$ becomes $rdrd\theta$ as follows.

Chop up the disk into grid corresponding to a grid size in polar coordinates of $\Delta r$ and $\Delta \theta$, just as we did for the intro reading. A single “curvy rectangle” with a corner $\left(r,\theta \right)$ is pictured below.

Let $\Delta A$ be area of the “curvy rectangle”. You can see that the area depends on the radius $r$, since its width is approximately $r\Delta \theta$ and its height is approximately $\Delta r$. Hence the area $\Delta A$ is approximately $r\Delta r\Delta \theta$. I hope that now you can believe that $dA$ becomes $rdrd\theta$ when changing variables to polar coordinates.

Example 2

Evaluate

$$\begin{array}{lll}\hfill {\iint }_D\left(x^2-y^2\right)dxdy& & \hfill \end{array}$$

where $D$ is the region pictured below.

Solution:

The region can be described simply if we change to coordinates $u$ and $v$ where

$$\begin{array}{llll}\hfill u& =y-x& \hfill & \\ \hfill v& =xy.& \hfill \text{ (1)}\end{array}$$

With this change of variables, our new region of integration $D^{\ast }$ is $0\le u\le 1$, $1\le v\le 2$.

Our change of variables as expressed in equation (1) give $u$ and $v$ in terms of $x$ and $y$. In our change of variables formula, we need to have $x$ and $y$ expressed in terms of $u$ and $v$ using some function $\left(x,y\right)=\mathbf{T}\left(u,v\right)$. So one way to solve this problem is to solve equation (1) for $x$ and $y$ to determine the function $\mathbf{T}$. We could then compute

$$\begin{array}{lll}\hfill \frac{\partial \left(x,y\right)}{\partial \left(u,v\right)},& & \hfill \end{array}$$

which is what we need for the change of variables formula.

For this example, we will use another method that skips this step. We will leave $u$ and $v$ expressed in terms of $x$ and $y$ (as in equation (1)) and compute $\frac{\partial \left(u,v\right)}{\partial \left(x,y\right)}$ instead of $\frac{\partial \left(x,y\right)}{\partial \left(u,v\right)}$. Since we know that

$$\begin{array}{lll}\hfill \frac{\partial \left(x,y\right)}{\partial \left(u,v\right)}=\frac{1}{\frac{\partial \left(u,v\right)}{\partial \left(x,y\right)}},& & \hfill \end{array}$$

we can easily get the correct factor for out change of variables formula.

To compute $\frac{\partial \left(u,v\right)}{\partial \left(x,y\right)}$, we view $u$ and $v$ as functions of $x$ and $y$: $s\left(x,y\right)$ and $t\left(x,y\right)$. The partial derivatives of $u$ and $v$ with respect to $x$ and $y$ are

$$\begin{array}{lll}\hfill \frac{\partial u}{\partial x}=-1,\frac{\partial u}{\partial y}=1,\frac{\partial v}{\partial x}=y,\text{ and}\frac{\partial v}{\partial y}=x.& & \hfill \end{array}$$

We compute

$$\begin{array}{llll}\hfill \frac{\partial \left(u,v\right)}{\partial \left(x,y\right)}& =det\left(\begin{array}{cc}\hfill \frac{\partial u}{\partial x}\hfill & \hfill \frac{\partial u}{\partial y}\hfill \\ \hfill \frac{\partial v}{\partial x}\hfill & \hfill \frac{\partial v}{\partial y}\hfill \end{array}\right)& \hfill & \\ \hfill & =\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial v}{\partial x}& \hfill & \\ \hfill & =-x-y,& \hfill & \end{array}$$

so that

$$\begin{array}{lll}\hfill \frac{\partial \left(x,y\right)}{\partial \left(u,v\right)}=\frac{1}{-x-y}.& & \hfill \end{array}$$

The factor we need for our change of variables formula is then

$$\begin{array}{lll}\hfill \left|\frac{\partial \left(x,y\right)}{\partial \left(u,v\right)}\right|=\frac{1}{x+y}& & \hfill \end{array}$$

since $x$ and $y$ are positive in our domain.

Once we change variables, we will need to integrate over the region $D^{\ast }$ the expression

$$\begin{array}{lll}\hfill \left(x^2-y^2\right)\frac{1}{x+y}.& & \hfill \end{array}$$

Since we can factor $x^2-y^2=\left(x-y\right)\left(x+y\right)$ the above expression simplifies to $x-y$.

Since the integral over $D^{\ast }$ is an integral in $u$ and $v$, we need to write $x-y$ in terms of $u$ and $v$. Glancing at equation (1), we see that $x-y=-u$, so we simply need to integrate $-u$ over the region $D^{\ast }$.

Our integral is

$$\begin{array}{llll}\hfill {\int \; <!--nolimits\; -->}_D\left(x^2-y^2\right)dxdy& ={\iint }_{D^{\ast }}-ududv& \hfill & \\ \hfill & =-{\int \; <!--nolimits\; -->}_1^2{\int \; <!--nolimits\; -->}_0^{1}ududv& \hfill & \\ \hfill & =-{\int \; <!--nolimits\; -->}_1^2{\left.\frac{u^2}{2}\right|}_0^{1}dv& \hfill & \\ \hfill & =-{\int \; <!--nolimits\; -->}_1^2\frac{1}{2}dv=-\frac{1}{2}& \hfill & \end{array}$$

where $x$ and $y$ are functions of $u$ and $v$.

Example 3

Evaluate

$$\begin{array}{lll}\hfill {\iint }_{D_c}{e}^{-x^2-y^2}dA& & \hfill \end{array}$$

where $D_c$ is the region in the first quadrant of the $xy$-plane where $x^2+y^2<c^2$.

Solution:

Change to polar coordinates. Region is sector: $0\le \theta \le \pi ∕2$ and $0\le r\le c$.

$$\begin{array}{llll}\hfill {\iint }_{D_c}{e}^{-x^2-y^2}dA& ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}{\int \; <!--nolimits\; -->}_0^c{e}^{-r^2}rdrd\theta & \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{\pi ∕2}{\left.-\frac{1}{2}{e}^{-r^2}\right|}_0^{c}d\theta & \hfill & \\ \hfill & =\frac{1}{2}\left(1-e^{-c^2}\right){\int \; <!--nolimits\; -->}_0^{2∕\pi }d\theta & \hfill & \\ \hfill & =\frac{\pi }{4}\left(1-e^{-c^2}\right)& \hfill & \end{array}$$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

3a. Double integrals

3b. Triple Integrals

3c. Changing variables

• Change of variables in double integrals
• Double integral change of variables examples
• Change of variables in triple integrals
• Triple integral change of variables examples

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material