Change of Variables

Theorem 5..4 By the one-to-one correspondence $\Phi(u,v) = (\phi(u,v), \psi(u,v))$ , the region $\Gamma$ is mapped into $\Omega$ . Furthermore, $\phi(u,v),\psi(u,v)$ are the class $C^{1}$ with respect to $u,v$

. Now suppose that Jacobian of $x,y$

with respect to $u,v$

$\displaystyle J(u,v) = \frac{\partial(x,y)}{\partial(u,v)} = \det\begin{pmatrix} \phi_{u}(u,v)&\phi_{v}(u,v)\\ \psi_{u}(u,v)&\psi_{v}(u,v) \end{pmatrix}$

is never be 0 on $\Gamma$ . Then the following is true for the continuous function $f(x,y)$

on $\Omega$ .

$\displaystyle \iint_{\Omega}f(x,y)dxdy = \iint_{\Gamma}f(\phi(u,v),\psi(u,v))\vert J(u,v)\vert dudv$

NOTE Let $\Omega$ be the region on $xy$ -plane and $\Gamma$ be the region on $uv$ -plane. Suppose that $\Phi$ is a map from $\Gamma$ to $\Omega$ satisfying

$\displaystyle x$	$\displaystyle =$	$\displaystyle \phi(u,v) = 2u$
$\displaystyle y$	$\displaystyle =$	$\displaystyle \psi(u,v) = 2v$

Then

$\displaystyle \binom{x}{y} = \left(\begin{array}{cc} 2 & 0\\ 0 & 2 \end{array} \right)\binom{u}{v} = \Phi\binom{u}{v}$

and $\Phi$ is invertible matrix. Thus there exists the inverse of $\Phi$ such that

$\displaystyle \binom{u}{v} = \Phi^{-1}\binom{x}{y} = \frac{1}{4}\left(\begin{array}{cc} 2 & 0\\ 0 & 2 \end{array} \right)\binom{x}{y}.$

Consider the rectangle with 4 vertices $(x,y), (x+\Delta x, y), (x, y + \Delta y), (x + \Delta x, y + \Delta y)$ .

**Figure 5.6:** $\Phi ^{-1}$
$\includegraphics[width = 9cm]{SOFTFIG-5/hensuhenkan.eps}$

Then the area of the rectangle is $\Delta x \Delta y$ . Now correspondence area of $uv$

-plane is given by $\displaystyle{\binom{u}{v} = \Phi^{-1}\binom{x}{y}}$

$\begin{displaymath}\begin{array}{l} (x,y) \to (\frac{x}{2},\frac{y}{2}) = (u,v)\... ...c{y + \Delta y}{2}) = (u + \Delta u, v + \Delta v) \end{array} \end{displaymath}$

Thus the area $\Delta u \Delta v$ of $uv$

-plane is

$\displaystyle \Delta u \Delta v$	$\displaystyle =$	$\displaystyle (u + \Delta u - u)(v + \Delta v - v)$
	$\displaystyle =$	$\displaystyle \left(\frac{x + \Delta x}{2} - \frac{x}{2}\right) \left(\frac{y + \Delta y}{2} - \frac{y}{2} \right) = \frac{1}{4} \Delta x \Delta y$

Thus

$\displaystyle \Delta x \Delta y = 4 \Delta u \Delta v = \vert\phi_u \psi_v - \phi_v \psi_u\vert\Delta u \Delta v = \vert J(u,v)\vert \Delta u \Delta v.$

This

is the jacobian.

Theorem 5..5 To transform the polar coordinate $(r,\theta)$ to the rectangular coordinate $(x,y)$

, since $x = r\cos{\theta}, y = r\sin{\theta}$ ,

$\displaystyle J(r,\theta) = \left\vert\begin{array}{cc} \frac{\partial x}{\part... ...s{\theta} \end{array}\right \vert = r(\cos^{2}{\theta} + \sin^{2}{\theta}) = r$

Thus

$\displaystyle \iint_{\Omega}f(x,y)dxdy = \iint_{\Gamma}f(r\cos{\theta},r\sin{\theta})r dr d\theta$

Example 5..4 Evaluate the following double integrals.
1. $\displaystyle{\iint_{\Omega}xydxdy, \Omega = \{(x,y): x^2 + y^2 \leq 1, x,y \geq 0\}}$
2. $\displaystyle{\iint_{\Omega}\frac{1}{x^2 + y^2}\:dxdy, \Omega = \{(x,y): 1 \leq x^2 + y^2 \leq 4, y \geq 0\}}$
3. $\displaystyle{\iint_{\Omega}(x-y)^2dxdy, \Omega = \{(x,y):\vert x+2y\vert \leq 1, \vert x-y\vert \leq1\}}$

SOLUTION 1. $\Omega$ is a circular region. Thus using polar coordinate, $x = r\cos{\theta}, y = r\sin{\theta}$ . Then $0 \leq x^2 + y^2 = r^2 \leq 1$ and $0 \leq r \leq 1$ . Also, $x = r\cos{\theta} \geq 0, y = r\sin{\theta} \geq 0$ and $\displaystyle{0 \leq \theta \leq \frac{\pi}{2}}$ . Thus $\Omega$ is transformed to

$\displaystyle \Gamma = \{(r,\theta) : 0 \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq 1 \}$

Then by Theorem5.5,

$\displaystyle \iint_{\Omega}xydxdy$	$\displaystyle =$	$\displaystyle \iint_{\Gamma}\underbrace{r\cos{\theta}}_{x}\underbrace{r\sin{\theta}}_{y}rdrd{\theta} = \iint_{\Gamma}r^3\cos{\theta}\sin{\theta}drd{\theta}$
	$\displaystyle =$	$\displaystyle \int_{0}^{\frac{\pi}{2}}\sin{\theta}\cos{\theta}d{\theta}\int_{0}... ...t \left[\frac{r^4}{4}\right ]_{0}^{1} = \frac{1}{8} \ensuremath{ \blacksquare}$

2. $\Omega$ is a washer region. Then by letting $x = r\cos{\theta}, y = r\sin{\theta}$ , we have $1 \leq x^2 + y^2 \leq 4$ and $1 \leq r^2 \leq4$ . Here since $r > 0$

, $1 \leq r \leq 2$ . Also since $y = r\sin{\theta} \geq 0$ , $0 \leq \theta \leq \pi$ . Thus $\Omega$ is transformed into

$\displaystyle \Gamma = \{(r,\theta) : 0 \leq \theta \leq \pi, 1 \leq r \leq 2\}$

$\displaystyle \iint_{\Omega} \frac{1}{x^2 + y^2} dxdy$	$\displaystyle =$	$\displaystyle \iint_{\Gamma}\frac{1}{r^2} \vert J(r,\theta)\vert dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi}\int_{1}^{2}\frac{1}{r^2} r dr d\theta = \int_{0}^{\pi}\int_{1}^{2}\frac{1}{r} dr d\theta$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi} d\theta [\log{r}]_{1}^{2} = \pi \log{2} \ensuremath{ \blacksquare}$

3. This double integral can be evaluated directly. But using the transformation of variables is easier.
Let $\left\{\begin{array}{l} x + 2y = u\\ x -y = v \end{array}\right.$ . Then solve for $x,y$

to get $\left\{\begin{array}{l} x = \frac{u+2v}{3}\\ y = \frac{u-v}{3} \end{array}\right.$ Substitute this into the condition of $\Omega$ . Then the point of $\Gamma = \{(u,v): \vert u\vert \leq 1, \vert v\vert \leq 1\}$ corresponds one-to-one into point in $\Omega$ . Now

$\displaystyle J(u,v) = \frac{\partial(x,y)}{\partial(u,v)} = \det\begin{pmatrix}1/3 & 2/3\ 1/3 & -1/3\end{pmatrix} = -\frac{1}{3}$

Thus,

$\displaystyle \iint_{\Omega}(x-y)^2 dx dy$	$\displaystyle =$	$\displaystyle \iint_{\Gamma}v^2 \vert J(u,v)\vert dudv$
	$\displaystyle =$	$\displaystyle \int_{-1}^1 du \int_{-1}^1 \frac{v^2}{3}dv = [u]_{-1}^{1}\left[\frac{v^3}{9}\right]_{-1}^{1}$
	$\displaystyle =$	$\displaystyle 2(\frac{2}{9}) = \frac{4}{9}\ensuremath{ \blacksquare}$

Exercise 5..4 Evaluate the following double integrals.
1. $\displaystyle{\iint_{\Omega}\sqrt{1 - x^2 - y^2}dxdy, \Omega = \{(x,y) : x^2 + y^2 \leq x\}}$
2. $\displaystyle{\iint_{\Omega}e^{(y-x)/(y+x)}dxdy, \Omega = \{(x,y) : x+y \leq 1, x \geq 0, y \geq 0\}}$

SOLUTION 1. $\Omega$ is a circlular region . Thus use the polar coordinate $x = r\cos{\theta}, y = r\sin{\theta}$ . Since $0 \leq x^2 + y^2 = r^2 \leq r\cos{\theta}$ , $r^2 \leq r\cos{\theta}$ and $r\cos{\theta} \geq 0$ . Now $r\cos{\theta} \geq 0$ implies $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$ . Since $0 \leq r^2 \leq r\cos{\theta}$ , $0 \leq r \leq \cos{\theta}$ . Thus, $\Omega$ is transformed into

$\displaystyle \Gamma = \{(r,\theta) : -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}, 0 \leq r \leq \cos{\theta} \}$

Then

$\displaystyle \iint_{\Omega}\sqrt{1 - x^2 - y^2}dxdy$

$\displaystyle =$

$\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{\cos{\theta}}\sqrt{1 - r^2}rdrd{\theta}$

Let

. Then $dt = -2r\:dr$ , $\begin{displaymath}\begin{array}{c\vert ccl} r & 0 & \to & \cos{\theta}\ \hline t & 1 & \to & 1-\cos^{2}{\theta} = \sin^{2}{\theta} \end{array}\end{displaymath}$ . Thus,

$\displaystyle \int_{0}^{\cos{\theta}}\sqrt{1 - r^2}rdr$	$\displaystyle =$	$\displaystyle -\frac{1}{2}\int_1^{\sin^{2}{\theta}}t^{\frac{1}{2}}\:dt = -\frac{1}{2}\cdot\frac{2}{3}\left[t^{3/2}\right]_1^{\sin^{2}{\theta}}$
	$\displaystyle =$	$\displaystyle -\frac{1}{3}(\sin^{3}{\theta} - 1) = \frac{1}{3}(1 - \sin^{3}{\theta})$

Thus,

$\displaystyle \iint_{\Omega}\sqrt{1 - x^2 - y^2}dxdy$

$\displaystyle =$

$\displaystyle \frac{1}{3}\int_{0}^{\frac{\pi}{2}}(1- \sin^{3}{\theta})d{\theta} = \frac{1}{3}(\frac{\pi}{2}-\frac{2}{3}) \ensuremath{ \blacksquare}$

2. Let

. Then check to see where the region $\Omega = \{(x,y) : x+y \leq1, x \geq 0, y \geq 0 \}$ map into.

Since

, the line $y = 0$

maps to

Since

, the line $x = 0$

maps to

A line

maps to

$\Omega$ is mapped to

$\displaystyle \Gamma = \{(u,v) : 0 \leq u \leq1, - u \leq v \leq u\}$

Thus

$\displaystyle I$	$\displaystyle =$	$\displaystyle \iint_{\Omega} e^{(y-x)/(y+x)} dxdy = \iint_{\Gamma} e^{v/u} \vert J(u,v)\vert du dv$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{1}\int_{u}^{u}e^{v/u}dvdu = \frac{1}{2}\int_{0}^{1}[ue^{v/u}]_{-u}^{u} du$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\int_{0}^{1}(ue - ue^{-1})du = \frac{1}{2}[\frac{u^2 e... ...frac{u^2}{2e}]_{0}^{1} = \frac{1}{4}(e-\frac{1}{e}) \ensuremath{ \blacksquare}$

Exercise A

1.

Evaluate the following double integrals．

(a) $\displaystyle{\iint_{\Omega}(x^2+y^{2})dxdy, \Omega = \{(x,y) : x^2 + y^2 \leq 4\}}$

(b) $\displaystyle{\iint_{\Omega}\frac{1}{(x^2+y^2)}dxdy, \Omega = \{(x,y) : 1 \leq x^2 + y^2 \leq 4, y \geq 0\}}$

(d) $\displaystyle{\iint_{\Omega}(x+y)dxdy, \Omega = \{(x,y): 0 \leq x+y \leq 1, \vert x-y\vert \leq 1}$

(e) $\displaystyle{\iint_{\Omega}\sqrt{4 - x^{2} - y^{2}}dxdy, \Omega = \{(x,y) : x^{2} + y^{2} \leq 2x \}}$

2.

Draw th graph of the region $\Gamma$ which is the image of $\Omega = \{(x,y): 0 \leq x + y \leq 1, 0 \leq x - y \leq 1\}$ by the transformation $u = x - y, v = x + y$

．Then using this transformation, evaluate $\iint_{\Omega}(2x + 3y)dxdy$ ．

3.

Evaluate $\iint_{\Omega}2xdxdy$ , where $\Omega = \{(x,y): 0 \leq x - y \leq 1, 0 \leq x + 2y \leq 1\}$ .

Exercise B

1.

Evaluate the following double integrals．

(a) $\displaystyle{\iint_{\Omega}x^{2}dxdy, \Omega = \{(x,y) : x^2 + y^2 \leq 4\}}$

(b) $\displaystyle{\iint_{\Omega}\log{(x^2+y^2)}dxdy, \Omega = \{(x,y) : 1 \leq x^2 + y^2 \leq 4\}}$

(d) $\displaystyle{\iint_{\Omega}e^{x^2 + y^2}dxdy, \Omega = \{(x,y) : 1 < x^2 + y^2 < 4 \}}$

(e) $\displaystyle{\iint_{\Omega}\sqrt{1 - x^2 - y^2}dxdy, \Omega = \{(x,y) : x^2 + y^2 \leq 1 \}}$

(f) $\displaystyle{\iint_{\Omega}(1 - x - 2y)dxdy, \Omega = \{(x,y) : x \geq 0, y \geq 0, x^2 + y^2 \leq 1 \}}$

2.

Using the tranformation $u = x + y, v = x - y$

, evaluate the following double integral.

$\displaystyle{\iint_{\Omega}(x^2 + y^2) e^{-x+y} dx dy, \Omega = \{-1 \leq x+y \leq 1, -1 \leq x - y \leq 1 \}}$