Change of variables examples

Example 1

Compute the double integral

$$\iint_{D}^{}$$

where g(x, y) = x² + y² 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

$$\theta \theta \theta$$

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

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

$$\iint_{D}^{} \iint_{{D^{\textstyle *}}}^{} \theta \left\vert\vphantom{ \det D{\mathbf{T}}(r,{\textstyle \theta})}\right. \theta \left.\vphantom{ \det D{\mathbf{T}}(r,{\textstyle \theta})}\right\vert \theta$$

where D^* is the region in the r$\theta$-plane that is mapped by 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^*.

We calculate

$$\theta \theta \theta \theta \theta$$

$${\frac{{\partial T_1}}{{\partial r}}} \theta \theta$$

$${\frac{{\partial T_2}}{{\partial r}}} \theta \theta$$

$${\frac{{\partial T_1}}{{\partial \theta}}} \theta \theta$$

$${\frac{{\partial T_2}}{{\partial \theta}}} \theta \theta$$

$$\theta \left\vert\vphantom{ \begin{array}{cc} \cos\theta & \sin\theta -r\sin \theta & r \cos\theta \end{array} }\right. \begin{array}{cc} \cos\theta & \sin\theta -r\sin \theta & r \cos\theta \end{array} \left.\vphantom{ \begin{array}{cc} \cos\theta & \sin\theta -r\sin \theta & r \cos\theta \end{array} }\right\vert$$

$$\theta \theta \theta \theta \theta$$

Since r$\ge$ 0 in D^*, we know that $\left\vert\vphantom{ \det D{\mathbf{T}}(r,{\textstyle \theta})}\right.$det DT(r,$\theta$)$\left.\vphantom{ \det D{\mathbf{T}}(r,{\textstyle \theta})}\right\vert$ = | r| = r. Therefore the integral is

$$\iint_{D}^{} \int_{0}^{{2\pi}} \int_{0}^{7} \theta$$

$$\int_{2}^{{2\pi}} {\frac{{r^4}}{{4}}} \bigg\vert _{0}^{7} \theta$$

$$\int_{2}^{{2\pi}} {\frac{{7^4}}{{4}}} \theta {\frac{{\pi 7^4}}{{2}}}$$

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

Chop up the disk into grid corresponding to a grid size in polar coordinates of $\Delta$r and $\Delta$$\theta$, just as we did for the intro reading. A single "curvy rectangle" with a corner (r,$\theta$) 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$. I hope that now you can believe that dA becomes r dr d$\theta$ when changing variables to polar coordinates.

Example 2

Evaluate

$$\iint_{D}^{}$$

where D is the region pictured below.

Solution:

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

u = y - x

v = xy. (1)

With this change of variables, our new region of integration D^* 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 (x, y) = T(u, v). So one way to solve this problem is to solve equation (1) for x and y to determine the function T. We could then compute

$${\frac{{\partial (x,y)}}{{\partial (u,v)}}}$$

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 $${\frac{{\partial (u,v)}}{{\partial (x,y)}}}$$ instead of $${\frac{{\partial (x,y)}}{{\partial (u,v)}}}$$. Since we know that

$${\frac{{\partial (x,y)}}{{\partial (u,v)}}} {\frac{{1}}{{\displaystyle \frac{\partial (u,v)}{\partial (x,y)}}}}$$

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

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

$${\frac{{\partial u}}{{\partial x}}} {\frac{{\partial u}}{{\partial y}}} {\frac{{\partial v}}{{\partial x}}} {\frac{{\partial v}}{{\partial y}}}$$

We compute

$${\frac{{\partial (u,v)}}{{\partial (x,y)}}} \left(\vphantom{ \begin{array}{cc} \displaystyle \frac{\partial u... ...{\partial x} & \displaystyle \frac{\partial v}{\partial y} \end{array} }\right. \begin{array}{cc} \displaystyle \frac{\partial u}{\partial x} & \... ...artial v}{\partial x} & \displaystyle \frac{\partial v}{\partial y} \end{array} \left.\vphantom{ \begin{array}{cc} \displaystyle \frac{\partial u... ...{\partial x} & \displaystyle \frac{\partial v}{\partial y} \end{array} }\right)$$

$${\frac{{\partial u}}{{\partial x}}} {\frac{{\partial v}}{{\partial y}}} {\frac{{\partial u}}{{\partial y}}} {\frac{{\partial v}}{{\partial x}}}$$ = - x - y,

so that

$${\frac{{\partial (x,y)}}{{\partial (u,v)}}} {\frac{{1}}{{-x -y}}}$$

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

$$\left\vert\vphantom{\frac{\partial (x,y)}{\partial (u,v)}}\right. {\frac{{\partial (x,y)}}{{\partial (u,v)}}} \left.\vphantom{\frac{\partial (x,y)}{\partial (u,v)}}\right\vert {\frac{{1}}{{x +y}}}$$

since x and y are positive in our domain.

Once we change variables, we will need to integrate over the region D^* the expression

$${\frac{{1}}{{x +y}}}$$

Since we can factor x² - y² = (x - y)(x + y) the above expression simplifies to x - y.

Since the integral over D^* 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^*.

Our integral is

$$\int_{D}^{} \iint_{{D^{\textstyle *}}}^{}$$

$$\int_{1}^{2} \int_{0}^{1}$$

$$\int_{1}^{2} {\frac{{u^2}}{{2}}} \bigg\vert _{0}^{1}$$

$$\int_{1}^{2} {\frac{{1}}{{2}}} {\frac{{1}}{{2}}}$$

where x and y are functions of u and v.

Example 3

Evaluate

$$\iint_{{D_c}}^{}$$

where D_c is the region in the first quadrant of the xy-plane where x² + y² < c².

Solution:

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

$$\iint_{{D_c}}^{} \int_{0}^{{\pi/2}} \int_{0}^{c} \theta$$

$$\int_{0}^{{\pi/2}} {\frac{{1}}{{2}}} \bigg\vert _{0}^{c} \theta$$

$${\frac{{1}}{{2}}} \left(\vphantom{1-e^{-c^2}}\right. \left.\vphantom{1-e^{-c^2}}\right) \int_{0}^{{2/\pi}} \theta$$

$${\frac{{\pi}}{{4}}} \left(\vphantom{1-e^{-c^2}}\right. \left.\vphantom{1-e^{-c^2}}\right)$$