Section 4.7 Change of Variables

4.7 Change of Variables

In previous sections, we use cylindrical or spherical coordinates to deal with some special integrations, so the the computation parts

could be easier to deal with. Recall that we set $x=r\text{cos}(\theta )$ and $y=r\text{sin}(\theta )$ such that $x$ and $y$ are function of two variables $r$,$\theta$, then we can set 

$$\int {\int }_{R}f(x,y)dA=\int {\int }_{S}f(r\text{cos}(\theta ),r\text{sin}(\theta ))rdrd\theta$$

where the region $R$ is replaced by the region $S$. We transfer the region $R$ into a new region $S$. We call this a transformation or mapping. We can do this in more general setting, not just for polar coordinates. 

Definition: Planar Transformations

A planar transformation $T$ is a function that transforms a region $G$ in one plane into a region $R$ in another plane by a change of variables. Both $G$ and $R$ are subsets of ${\mathbb{R}}^2$. A transformation $T$: $G\to R$, defined as $T(u,v)=(x,y)$, is said to be a one-to-one transformation if no two points map to the same image point.

Example 1: Suppose a transformation $T$ is defined as $T(r,\theta )=(x,y)$ where $x=r\text{cos}(\theta )$, $y=r\text{sin}(\theta )$. Find the image of the polar rectangle $G=\{(r,\theta )|1<r\le 2,\frac{\pi }{2}\le \theta \le \pi \}$ in the $r\theta$-plane to a region $R$ in the $xy$-plane.

Exercise 1: Suppose a transformation $T$ is defined as $T(r,\theta )=(x,y)$ where $x=r\text{cos}(\theta )$, $y=r\text{sin}(\theta )$. Find the image of the polar rectangle $G=\{(r,\theta )|0<r\le 1,0\le \theta \le \frac{\pi }{2}\}$ in the $r\theta$-plane to a region $R$ in the $xy$-plane.

Example 2: Find the image of $S$ under the transformation where $S$ is square bounded by $u=0$, $u=1$, $v=0$, $v=1$, $x=u$, $y=v(1+u^2)$. Draw both regions. 

Exercise 2: Find the image of $S$ under the transformation where $S$ is square bounded by $u=0$, $u=1$, $v=0$, $v=1$, $x=v$, $y=u(1+v^2)$. Draw both regions. 

When we do the double integration after change the variables $x=r\text{cos}(\theta )$ and $y=r\text{sin}(\theta )$, we have to change $dA$ into $rdrd\theta$, the reason is that when we do Riemann sum, we have $dA=rdrd\theta$. At here $r$ is actually called Jacobian. 

Definition

The Jacobian of the $C^1$ transformation $T(u,v)=(g(u,v),h(u,v))$ is denoted by $J(u,v)$ and is defined by the $2\times 2$ determinant

$$J(u,v)=\frac{\partial (x,y)}{\partial (u,v)}=\left|\begin{array}{ccc}\frac{\partial x}{\partial u}& \frac{\partial y}{\partial u}\frac{\partial x}{\partial v}& \frac{\partial y}{\partial v}\end{array}\right|=(\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}).$$

Reasoning: Use Cross product and the area of parallelogram. Show the case of $x=r\text{cos}(\theta )$ and $y=r\text{sin}(\theta )$.

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 $C^1$ 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 

$$\int {\int }_{R}f(x,y)dA=\int {\int }_{S}f(g(u,v),h(u,v))\left|\frac{\partial (x,y)}{\partial (u,v)}\right|dudv.$$

Example 3: Find Jacobian where $x=u+u^{2}v$ and $y=uv$.

Exercise 3: Find Jacobian where $x=u^{2}v$ and $y=u+v^2$. 

Example 4: Use transformation, $x=2u+v$ and $y=u-v$ to evaluate the integral $\int {\int }_R(x+y)dA$ where $R$ is the region with vertices $(0,0),(1,-1)$, $(2,1)$ and $(3,0)$.

Exercise 4: Use transformation, $x=u-v$ and $y=u+v$ to evaluate the integral $\int {\int }_R(x-y)dA$ where $R$ is the region with vertices $(0,0),(1,1)$, $(-1,1)$, and $(0,2)$.

Example 5: Find equations of a transformation $T$ that maps a rectangular region $S$ in the $uv$-plane onto $R$, the parallelogram with vertices $(0,0)$, $(2,4),(3,3)$ and $(1,-1)$.

Exercise 5: Find equations of a transformation $T$ that maps a rectangular region $S$ in the $uv$-plane onto $R$, the parallelogram with vertices $(0,0)$, $(1,0),(2,1)$ and $(1,1)$.

Example 6: Evaluate using change of variables $\int {\int }_R\frac{x-2y}{3x-y}dA$ where $R$ is the parallelogram enclosed by the lines $x-2y=0$, $x-2y=4$, $3x-y=1$ and $3x-y=8$.

Exercise 6: Evaluate using change of variables $\int {\int }_R(2x+y)(x-3y)dA$ where $R$ is the parallelogram enclosed by the lines $2x+y=0$, $2x+y=3$, $x-3y=-1$ and $x-3y=2$.

Example 7: Evaluate the integral using transformation $\int {\int }_R(x-y)dA$ where $R$ is the parallelogram with vertices $(0,0)$, $(2,4),(3,3)$ and $(1,-1)$.

Group work:

1. Evaluate the integral using transformation $\int {\int }_R(x+y)dA$ where $R$ is the parallelogram with vertices $(0,0)$, $(1,0),(2,1)$ and $(1,1)$.

2. Evaluate using change of variables $\int {\int }_R(x+y)e^{x^2-y^2}dA$ where $R$ is the parallelogram enclosed by the lines $x-y=0$, $x-y=1$, $x+y=1$ and $x+y=2$.

3. Evaluate the integral using transformation $\int {\int }_R(x+2y)dA$ where $R$ is the parallelogram with vertices $(0,0)$, $(4,3),(2,4)$ and $(-2,1)$.