Math 344: Calculus III

15.9 Change of Variables

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15.9.1 Revisiting Polar Coordinates

We already learned the usefulness of polar coordinates when integrating over a domain with circular symmetry. The new coordinate system $\langle r,\theta \rangle$ are related to the old coordinates $\langle x,y \rangle$ by the system of equations

$$ \require{color} \definecolor{brightblue}{rgb}{.267, .298, .812} \definecolor{darkblue}{rgb}{.08, .18, .28} \definecolor{palepink}{rgb}{1, .73, .8} \definecolor{softmagenta}{rgb}{.99,.34,.86} \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \newcommand{\pypx}[2][x]{\dfrac{\partial #2}{\partial #1}} \newcommand{\dydx}[2][x]{\dfrac{d #2}{d #1}} \newcommand{\deltax}[2][x]{\frac{\Delta #2}{\Delta #1}} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} r\cos(\theta) \\ r\sin(\theta) \end{bmatrix} = T\left(\begin{bmatrix} r \\ \theta \end{bmatrix}\right) $$

If $S$ is a circular domain in the $r\theta$-plane so that $T(S) = R$, a circular domain in the $xy$-plane then

Figure 1

This map $T:\mathbb{R}^2\mapsto\mathbb{R}^2$ is a bijection . This means

1. Every point $\langle x,y \rangle$ is the image of only one point $\langle r,\theta \rangle$.
   
2. The map $T$ is an invertible function.

In this case the domains $R$ and $S$ are identical. In fact the domain and co-domain of $T$ differ only by the coordinates used to describe each point on the plane $\mathbb{R}^2$, not the plane itself. These properties are very important when we want to use such a change of variables map in integration. If we chose a map that is not invertible, that is, not a bijection from $S$ to $R$, then the integral over $S$ will not give us the correct result.

Moreover we usually require that the map $T$ have both first order partial derivatives $\frac{\partial T}{\partial r}$ and $\frac{\partial T}{\partial \theta}$, and that these partial derivatives must be continuous . This allows us to easily compute the Jacobian. When a function has all possible first order derivatives and these first order partial derivatives are continuous, we say that the map $T$ is a $\mathbf{C^1}$ transformation .

• A continuous function $f$ on domain $R$ is called $C^0(R)$, or $f\in C^0(R)$.
  
• A function with continuous first order partial derivatives for every input variable in domain $R$ is called $C^1(R)$, or $f\in C^1(R)$.
  
• A function with continuous mixed second order partial derivatives for every combination of input variables in domain $R$ is called $C^2(R)$, or $f\in C^2(R)$.

$$ \pypx[r]{T} = \begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix} \qquad \pypx[\theta]{T} = \begin{bmatrix} -r\sin(\theta) \\ \ \ r\cos(\theta) \end{bmatrix} $$
Hence our map $T:\mathbf{R}^2\mapsto\mathbf{R}^2$ is $C^1$, or $T\in C^1(R)$.

$$ \begin{array}{ccc} \dfrac{\partial^2 T}{\partial r^2} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} & \qquad & \dfrac{\partial^2 T}{\partial\theta\,\partial r} = \begin{bmatrix} -\sin(\theta) \\ \ \ \cos(\theta) \end{bmatrix} \\ \\ \dfrac{\partial^2 T}{\partial r\,\partial\theta} = \begin{bmatrix} -\sin(\theta) \\ \ \ \cos(\theta) \end{bmatrix} & \qquad & \dfrac{\partial^2 T}{\partial\theta^2} = \begin{bmatrix} -r\cos(\theta) \\ -r\sin(\theta) \end{bmatrix} \end{array} $$
Since all four of our mixed second partial derivatives are continuous, map $T$ is $C^2$, or $T\in C^2(R)$.

In fact we can differential polynomials, $\cos$, and $\sin$ as many times as we want so we say that function $T$ is $C^{\infty}$, or $T\in C^{\infty}(R)$. Of course computing $n^2$ mixed partial derivatives of order $n$ would be very tedious. It does not help that all mixed partial derivatives of order greater than 1 with respect to $r$ would be zero.

Fortunately we only need our maps to be $C^1$ so that we can create the Jacobian matrix

$$ \begin{bmatrix} \dfrac{\partial T}{\partial r} & \dfrac{\partial T}{\partial\theta} \end{bmatrix} = \begin{bmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial y}{\partial r} \\ \dfrac{\partial x}{\partial\theta} & \dfrac{\partial y}{\partial\theta} \end{bmatrix} $$
We learned in Section 15.3 that while small differences in each coordinate are given by

$$ \begin{align*} \Delta x &= x_1 - x_0 = r_1\cos(\theta_1) - r_0\cos(\theta_0) \\ \Delta y &= y_1 - y_0 = r_1\sin(\theta_1) - r_0\sin(\theta_0) \end{align*} $$
small differences in area are computed

$$ \Delta A = \Delta x\Delta y = {\color{magenta} r}\Delta r\Delta\theta $$
where $r_0\le {\color{magenta} r}\le r_1$.

This factor shows up in our double integral by use of Riemann sums

$$ \begin{align*} \displaystyle\iint_D f(x,y)\,dx\,dy &= \displaystyle\lim_{m\rightarrow\infty}\displaystyle\lim_{n\rightarrow\infty} \displaystyle\sum_{j=0}^m\displaystyle\sum_{k=0}^n f(x_j^*,y_k^*)\,\Delta A \\ &= \displaystyle\lim_{m\rightarrow\infty}\displaystyle\lim_{n\rightarrow\infty} \displaystyle\sum_{j=0}^m\displaystyle\sum_{k=0}^n f(r_j^*\cos(\theta_j^*),r_k^*\sin(\theta_k^*))\,{\color{magenta} r}\,dr\,d\theta \\ &= \displaystyle\iint_D f(r\cos(\theta),r\sin(\theta))\,{\color{magenta} r}\,dr\,d\theta \end{align*} $$
The factor $\color{magenta}r$ is called the Jacobian . The Jacobian for rectangular coordinates is $1$ since

$$ \Delta A = 1\cdot\Delta x\Delta y $$

Example 1

Consider the function $f:\mathbb{R}^2\mapsto\mathbb{R}$ given by $f(x,y) = 1 - x^2 - y^2$. In polar coordinates we can describe this function

$$ f(x,y) = 1 - x^2 - y^2 = 1 - r^2 = f(r\cos(\theta), r\sin(\theta)) = g(r,\theta) $$
Here function $g = f$ for every vector $\langle x, y\rangle$ in $\mathbb{R}^2$. The change in coordinates is the only difference between these two maps $f$ and $g$ that represent the same function with respect to different coordinate systems. Recall from Section 14.4 that we derived the equation for the tangent plane

$$ \Delta z = \frac{\partial z}{\partial x}\Delta x + \frac{\partial z}{\partial y}\Delta y $$
or

$$ \begin{align*} \Delta x &= \frac{\partial x}{\partial r}\,\Delta r + \frac{\partial x}{\partial\theta}\,\Delta\theta \\ \Delta y &= \frac{\partial y}{\partial r}\,\Delta r + \frac{\partial y}{\partial\theta}\,\Delta\theta \end{align*} $$
From this we obtain the linear equation

$$ \begin{bmatrix} \Delta x \\ \Delta y \end{bmatrix} = \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial\theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial\theta} \end{bmatrix}\,\begin{bmatrix} \Delta r \\ \Delta\theta \end{bmatrix} $$
or

$$ \begin{bmatrix} \frac{\partial z}{\partial x} \\ \frac{\partial z}{\partial y} \end{bmatrix} = \begin{bmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial\theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial\theta} \end{bmatrix}\,\begin{bmatrix} \frac{\partial z}{dr} \\ \frac{\partial z}{\partial \theta} \end{bmatrix} $$
In linear algebra this is the transition matrix from one coordinate system in the tangent space with basis vectors $\left\{ \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} \right\}$ to the tangent space with basis vectors $\left\{ \frac{\partial z}{\partial\theta}, \frac{\partial z}{\partial r} \right\}$. The determinate of matrix $S$ is defined to be the change in scale introduced by this change of coordinates .

If we have a region $R$ in the tangent space with area $A$ described in rectangular coordinates, then the same area would be det$(S)$ times the area computed using equivalent polar coordinates.

Figure 2

Thus we have

$$ \Delta A = \Delta x\,\Delta y = {\color{magenta} r}\,\Delta\theta\,\Delta r $$

This scaling factor is called the Jacobian the determinant of the transition matrix $S$. Recall that the determinant is signed .

• A positive determinant means that the orientation of the surface is unchanged.
  
• A negative determinant means that the orientation of the surface has been reversed or flipped .

Computing the determinate we have that the Jacobian $J$ is given by

$$ \begin{align*} J &= \left|S\right| = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial\theta} & \frac{\partial y}{\partial\theta} \end{vmatrix} \\ \\ &= \begin{vmatrix} \cos(\theta) & \sin(\theta) \\ -r\sin(\theta) & r\cos(\theta) \end{vmatrix} \\ \\ &= r\cos^2(\theta) + r\sin^2(\theta) = r\\ \end{align*} $$
For our double integral, we do not need the sign of the determinant. In fact we are only concerned with the scaling factor, not the orientation. Hence we have that the scaling factor using in our double integral is the absolute value of the Jacobian. Since in polar coordinates the magnitude $r$ is always greater than or equal to zero we have $|r| = r$ and

$$ \displaystyle\iint_D f(x,y)\,dA = \displaystyle\iint_D f(r\cos(\theta),r\sin(\theta)\,r\,dr\,d\theta $$
We denote the Jacobian using similar to the chain rule to remind us of the similarity

$$ J = \frac{\partial(x,y)}{\partial(r,\theta)} $$
so that

$$ \begin{align*} \displaystyle\iint_D f(x,y)\,dA &= \displaystyle\iint_D f(r\cos(\theta),r\sin(\theta)\,\left|J\right|\,dr\,d\theta \\ \\ &= \displaystyle\iint_D f(r\cos(\theta),r\sin(\theta)\,\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right|\,dr\,d\theta \end{align*} $$

15.9.2 Revisiting Cylindrical Coordinates

In the same way we can compute the Jacobian scaling factor in our change of coordinates from rectangular coordinates $\langle\,x,y,z\,\rangle$ to cylindrical coordinates $\langle\,r,\theta,z\,\rangle$ as follows.

$$ \begin{align*} J &= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z} \\ \end{vmatrix} = \begin{vmatrix} \cos(\theta) & -r\sin(\theta) & 0 \\ \sin(\theta) & r\cos(\theta) & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ \\ &= 0 - 0 + 1\cdot\begin{vmatrix} \cos(\theta) & -r\sin(\theta) \\ \sin(\theta) & r\cos(\theta) \end{vmatrix} \\ \\ &= r\cos^2(\theta) + r\sin^2(\theta) = r \end{align*} $$
In the triple integral we have

$$ |J| = \left|\frac{\partial(x,y,z)}{\partial(r,\theta,z)}\right| = r $$
since in cylindrical coordinates $r\ge 0$.

15.9.3 Revisiting Spherical Coordinates

Likewise given the change of variables from rectangular coordinates $\langle\,x,y,z\,\rangle$ to spherical coordinates $\langle\,\rho,\theta,\phi\,\rangle$.

$$ \begin{align*} x &= \rho\cos(\theta)\sin(\phi) \\ y &= \rho\sin(\theta)\sin(\phi) \\ z &= \rho\cos(\phi) \end{align*} $$
We may compute the Jacobian

$$ \begin{align*} J &= \begin{vmatrix} \frac{\partial x}{\partial\rho} & \frac{\partial x}{\partial\theta} & \frac{\partial x}{\partial\phi} \\ \frac{\partial y}{\partial\rho} & \frac{\partial y}{\partial\theta} & \frac{\partial y}{\partial\phi} \\ \frac{\partial z}{\partial\rho} & \frac{\partial z}{\partial\theta} & \frac{\partial z}{\partial\phi} \end{vmatrix} = \begin{vmatrix} \cos(\theta)\sin(\phi) & -\rho\sin(\theta)\sin(\phi) & \rho\cos(\theta)\cos(\phi) \\ \sin(\theta)\sin(\phi) & \rho\cos(\theta)\sin(\phi) & \rho\sin(\theta)\cos(\phi) \\ \cos(\phi) & 0 & -\rho\sin(\phi) \end{vmatrix} \\ \\ &= \cos(\phi)\begin{vmatrix} -\rho\sin(\theta)\sin(\phi) & \rho\cos(\theta)\cos(\phi) \\ \rho\cos(\theta)\sin(\phi) & \rho\sin(\theta)\cos(\phi) \end{vmatrix} - 0 \\ \\ &\ \qquad\qquad\qquad - \rho\sin(\phi)\begin{vmatrix} \cos(\theta)\sin(\phi) & -\rho\sin(\theta)\sin(\phi) \\ \sin(\theta)\sin(\phi) & \rho\cos(\theta)\sin(\phi) \end{vmatrix} \\ \\ &= \cos(\phi)\left(-\rho^2\sin^2(\theta)\sin(\phi)\cos(\phi) - \rho^2\cos^2(\theta)\sin(\phi)\cos(\phi) \right) \\ \\ &\ \qquad\qquad\qquad - \rho\sin(\phi)\left(\rho\cos^2(\theta)\sin^2(\phi) + \rho\sin^2(\theta)\sin^2(\phi) \right) \\ \\ &= -\rho^2\sin(\phi)\cos^2(\phi)\left(\sin^2(\theta) + \cos^2(\theta) \right) -\rho^2\sin^3(\phi)\left( \cos^2(\theta) + \sin^2(\theta) \right) \\ \\ &= -\rho^2\sin(\phi)\cos^2(\phi) - \rho^2\sin^3(\phi) = -\rho^2\sin(\phi)\left( \cos^2(\phi) + \sin^2(\phi) \right) \\ \\ &= -\rho^2\sin(\phi) \end{align*} $$

In our triple integral, we utilize the absolute value of the Jacobian

$$ |J| = \left|\frac{\partial(x,y,z)}{\partial(\rho,\theta,\phi)}\right| = \rho^2\sin(\phi) $$
The last equality because $\rho^2\ge 0$ and the sine function is also nonnegative on the interval $[0,\pi]$.

15.9.4 Change of Variables in a Double Integral

Definition

Suppose that $T$ is a $C^1$ transformation given by

$$ \begin{align*} x &= g(u,v) \\ y &= h(u,v) \end{align*} $$
Then we denote the Jacobian $J = \frac{\partial(x,y)}{\partial(u,v)}$ and

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

A good theorem for using the Jacobian requires several assumptions about the change of variables function $T:R\rightarrow S$ and the geometry of the domain and range of $T$.

• Both $R$ and $S$ must be type I or type II plane regions.
• T must be a bijection from $R$ to $S$ except possibly on the boundary of $S$.
• $T$ must be a $C^1$ function so that we can compute the Jacobian.
• The Jacobian of $T$ must be nonzero on all of $S$.

The geometry of the regions $R$ and $S$ are of equal importance to the properties of $T$.

Theorem 1

Suppose that $T\in C^1(R)$ is a bijection from region $R$ to region $S$ such that regions $R$ and $S$ are of type I or type II plane regions. Suppose $T:R(x,y)\rightarrow S(u,v)$ defined by

$$ \begin{align*} x &= x(u,v) \\ y &= y(u,v) \end{align*} $$
is a bijection except possibly on the boundary of $S$, whose Jacobian

$$ J = \frac{\partial(x,y)}{\partial(u,v)} $$
is nonzero on $S$. Suppose also that function $f:R\rightarrow\mathbb{R}$ is continuous on $R$. Then

$$ \displaystyle\iint_S f(x,y)\,dA = \displaystyle\iint_S f(g(u,v),h(u,v))\,\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv $$

15.9.5 Additional Practice

Example 1

Compute the integral

$$ \displaystyle\int_0^3\displaystyle\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\displaystyle\int_0^4 \sqrt{x^2+y^2}\,dz\,dy\,dx $$

Follow Along

Draw a picture! The limits of integration indicate that the interval for $x$ is $[0,3]$ and $z$ is $[0,4]$. However the limits for $y$ are given by
$$ \begin{align*} -\sqrt{9-x^2} &= y = \sqrt{9-x^2} \\ |y| &= \sqrt{9-x^2} \\ y^2 &= 9 - x^2 \\ x^2 + y^2 &= 9 \end{align*} $$
Figure 3 In the $xy$-plane the cross section is a semi-circle so we will change variables to cylindrical coordinates,
$$ \begin{align*} \displaystyle\int_0^3\displaystyle\int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}\displaystyle\int_0^4 \sqrt{x^2+y^2}\,dz\,dy\,dx &= \displaystyle\int_0^4\displaystyle\int_0^{\pi}\displaystyle\int_0^3 r\,r\,dr\,d\theta\,dz \\ \\ &= \displaystyle\int_0^4\displaystyle\int_0^{\pi}\left[\displaystyle\int_0^3 r^2\,dr\right]\,d\theta\,dz \\ \\ &= \displaystyle\int_0^4\displaystyle\int_0^{\pi}\left[\frac{r^3}{3}\right]_0^3\,d\theta\,dz \\ \\ &= 9\left[\displaystyle\int_0^4\,dz\right]\cdot\left[\displaystyle\int_0^{\pi}\,d\theta\right] \\ \\ &= 9(4)(\pi) = 36\pi \end{align*} $$

Example 2

Find the mass $m$ and center of mass $(\bar{x},\bar{y},\bar{x})$ of a solid $S$ in the first octant bounded by the coordinate planes, the plane $z=4$, and the cylinder $x^2 + y^2 = 16$, if its density at a point $P$ is proportional to the distance of $P$ from the $z$-axis.

Follow Along

Draw a picture! Figure 4
$$ \begin{align*} m &= \displaystyle\int_0^4\displaystyle\int_0^4\displaystyle\int_0^{\sqrt{4-x^2}} \sigma(x,y,z)\,dy\,dx\,dz \\ \\ &=\displaystyle\int_0^4\displaystyle\int_0^{\pi/2}\displaystyle\int_0^4 kr\,r\,dr\,d\theta\,dz \\ \\ &= \left[\displaystyle\int_0^4 \,dz\right]\cdot\left[\displaystyle\int_0^{\pi/2}\,d\theta\right]\cdot\left[\displaystyle\int_0^4 kr^2\,dr\right] \\ \\ &= \frac{\pi}{2}\left(4\right)\left[\frac{kr^3}{3}\right]_0^4 = 2k\pi\left(\frac{64}{3} \right) = \frac{128k\pi}{3} \end{align*} $$
Now $\bar{x}$ is the mean distance from the $yz$-plane, so
$$ \begin{align*} M_{yz} &= \displaystyle\int_0^4\displaystyle\int_0^4\displaystyle\int_0^{\sqrt{4-x^2}} {\color{magenta}x}\,\sigma(x,y,z)\,dy\,dx\,dz \\ \\ &= \displaystyle\int_0^4\displaystyle\int_0^{\pi/2}\displaystyle\int_0^4 {\color{magenta}r\cos(\theta)}\,kr\,r\,dr\,d\theta\,dz \\ \\ &= k\,\left[\displaystyle\int_0^4\,dz\right]\cdot\left[\displaystyle\int_0^{\pi/2}\cos(\theta)\,d\theta\right]\cdot\left[\displaystyle\int_0^4 r^3\,dr\right] \\ \\ &= k\,\left(4\right)\cdot\left(1\right)\cdot\left[\frac{r^4}{4}\right]_0^4 = 4k\left(64\right) = 256k \\ \end{align*} $$
Similarly $\bar{y}$ is the mean distance from the $xz$-plane,
$$ \begin{align*} M_{xz} &= \displaystyle\int_0^4\displaystyle\int_0^4\displaystyle\int_0^{\sqrt{4-x^2}} {\color{magenta}y}\,\sigma(x,y,z)\,dy\,dx\,dz \\ \\ &= \displaystyle\int_0^4\displaystyle\int_0^{\pi/2}\displaystyle\int_0^4 {\color{magenta}r\sin(\theta)}\,kr\,r\,dr\,d\theta\,dz \\ \\ &= k\,\left[\displaystyle\int_0^4\,dz\right]\cdot\left[\displaystyle\int_0^{\pi/2}\sin(\theta)\,d\theta\right]\cdot\left[\displaystyle\int_0^4 r^3\,dr\right] \\ \\ &= k\,\left(4\right)\cdot\left(1\right)\cdot\left[\frac{r^4}{4}\right]_0^4 = 4k\left(64\right) = 256k \\ \\ \end{align*} $$
Finally $\bar{z}$ is the mean distance from the $xy$-plane,
$$ \begin{align*} M_{xy} &= \displaystyle\int_0^4\displaystyle\int_0^4\displaystyle\int_0^{\sqrt{4-x^2}} {\color{magenta}z}\,\sigma(x,y,z)\,dy\,dx\,dz \\ \\ &= \displaystyle\int_0^4\displaystyle\int_0^{\pi/2}\displaystyle\int_0^4 {\color{magenta}z}\,kr\,r\,dr\,d\theta\,dz \\ \\ &= k\,\left[\displaystyle\int_0^4 z\,dz\right]\cdot\left[\displaystyle\int_0^{\pi/2}\,d\theta\right]\cdot\left[\displaystyle\int_0^4 r^2\,dr\right] \\ \\ &= k\,\left[\frac{z^2}{2} \right]_0^4\cdot\left(\frac{\pi}{2}\right)\cdot\left[\frac{r^3}{3}\right]_0^4 \\ \\ &= k\,\left(8\right)\cdot\left(\frac{\pi}{2}\right)\cdot\left(\frac{64}{3}\right) = \frac{256k\pi}{3} \end{align*} $$
Hence we have
$$ \begin{align*} \bar{x} &= \frac{M_{yz}}{m} = 256k\dfrac{3}{128k\pi} = \frac{6}{\pi} \\ \\ \bar{y} &= \frac{M_{xz}}{m} = 256k\dfrac{3}{128k\pi} = \frac{6}{\pi} \\ \\ \bar{z} &= \frac{M_{xy}}{m} = \frac{256k\pi}{3}\frac{3}{128k\pi} = 2 \\ \\ \left(\bar{x},\bar{y},\bar{z}\right) &= \left(\frac{6}{\pi}, \frac{6}{\pi}, 2\right) \end{align*} $$

Example 3

Evaluate the integral

$$ \displaystyle\iint_R (4x + 8y)\,dA $$
where $R$ is the parallelogram with vertices $(-1,3)$, $(1,-3)$, $(3,-1)$, and $(1,5)$.

Follow Along

Draw a picture! The limits of integration indicate that the interval for $z$ is $[0,4]$. However the limits for $x$ and $y$ are more complicated. Figure 5 The left figure shows the graph of $f(x,y) = 4x + 8y$ over the domain. The right figure shows the domain only. To compute the double integral one could divide the domain into two sub-domains. Figure 6 For sub-domains $D_1$ and $D_2$, one may compute the values for $y$ in terms of $x$ using the equations of the line segments that make up the boundary of the domain. This technique will work however one may consider a __change of variables__.
$$ \begin{align*} x &= \frac{1}{4}(u + v) \\ y &= \frac{1}{4}(v - 3u) \end{align*} $$ Figure 7 This constitutes a bijective, $C^1$ map $T:S\mapsto R$ so that we can compute new limits and the Jacobian. We compute the map $T^{-1}$ so that we can determine the limits of integration,

$ \ \ \ \ \ \ \ \ \ \ \ \ $	$\ \ \ $System For $u$	$\ \ \ $System For $v$	$ \ \ \ \ \ \ \ \ \ \ \ \ $
$ \ \ \ \ \ \ \ \ \ \ \ \ $	$\begin{align*} \large{4x} &\large{= u + v} \\ \large{-4y} &\large{= 3u - v} \\ \hline \large{4x - 4y} &\large{= 4u} \end{align*}$	$\begin{align*} \large{12x} &\large{= 3u + 3v} \\ \large{4y} &\large{= -3u + v} \\ \hline \large{12x + 4y} &\large{= 4v} \end{align*}$	$ \ \ \ \ \ \ \ \ \ \ \ \ $

$$ \begin{align*} u &= x - y \\ v &= 3x + y \\ \end{align*} $$

$$ \begin{align*} u(-1,3) &= -4 \qquad &u(1,-3) &= 4 \qquad &u(3,-1) &= 4 \qquad &u(1,5) &= -4 \\ v(-1,3) &= 0 \qquad &v(1,-3) &= 0 \qquad &u(3,-1) &= 8 \qquad &v(1,5) &= 8 \end{align*} $$
The Jacobian is given by
$$ \begin{align*} J &= \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{1}{4} & -\frac{3}{4} \\ \frac{1}{4} &\ \ \frac{1}{4} \end{vmatrix} = \dfrac{1}{16} + \dfrac{3}{16} = \dfrac{1}{4} \end{align*} $$
Now we can compute the integral
$$ \begin{align*} \displaystyle\iint_R (4x + 8y)\,dA &= \displaystyle\iint_S (4x(u,v) + 8y(u,v))\,|J|\,du\,dv \\ \\ &= \displaystyle\int_0^8 \displaystyle\int_{-4}^4 (u + v) + 2(v - 3u)\,\dfrac{1}{4}\,du\,dv \\ \\ &= \dfrac{1}{4} \displaystyle\int_0^8\,\left[\,\displaystyle\int_{-4}^4 3v - 5u\,du\,\right]\,dv \\ \\ &= \dfrac{1}{4} \displaystyle\int_0^8\,\left[\,3uv - \frac{5}{2}u^2\,\right]_{-4}^4\,dv \\ \\ &= \dfrac{1}{4} \displaystyle\int_0^8\,\left(12v - 40 - \left(-12v - 40\right)\right)\,dv \\ \\ &= \dfrac{1}{4} \displaystyle\int_0^8 24v\,dv \\ \\ &= \dfrac{1}{4} \left[ 12v^2 \right]_0^8 = 3(64) = 192 \end{align*} $$

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