For any smooth transformation \(x(u,v),y(u,v),\) the double integral over a region \(S\) is given by
\begin{equation*}
\dint_S f(x(u,v),y(u,v))\left| \begin{array}{rr}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\
\end{array} \right| \: du \: dv
\end{equation*}
Example 4.4.5.
Integrate \(f(x,y)=3x+4y\) over the region with vertices with coordinates (2,1), (10,5), (13,14), and (5,10) using the transformations \(u(x,y)=\frac{3}{5}x-\frac{1}{5}y\) and \(v(x,y)=-\frac{1}{5}x+\frac{2}{5}y\text{.}\)
For the Jacobian and the substitution we need to have \(x(u,v)\) and \(y(u,v)\text{.}\) Our first step is to invert the transformation provided. We do this by solving the system of equations for \(x\) and \(y\text{.}\) We can do this by our favorite method of solving systems of linear equations (Gaussian elimination or substitution both work). The result is \(x(u,v)=2u+v\) and \(y(u,v)=u+3v\text{.}\)
Our second step is to calculate the Jacobian.
\begin{align*}
\frac{\partial x}{\partial u} = & 2.\\
\frac{\partial x}{\partial v} = & 1.\\
\frac{\partial y}{\partial u} = & 1.\\
\frac{\partial y}{\partial v} = & 3.\\
\left| \begin{array}{rr}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \\
\end{array} \right| = & (2)(3)-(1)(1)\\
= & 5.
\end{align*}
Our third step is to substitute the transformation into the function.
\begin{align*}
f(x(u,v),y(u,v)) = & 3(2u+v)+4(u+3v) \\
= & 10u+15v.
\end{align*}
The fourth step before evaluating the integral is to find the region over which we are integrating. For this we must transform each of the vertices of the region which is given in terms of \(u\) and \(v\text{.}\)
\begin{align*}
u(2,1) = & \frac{3}{5}(2)-\frac{1}{5}(1) & = 1.\\
v(2,1) = & -\frac{1}{5}(2)+\frac{2}{5}(1) & = 0.\\
u(10,5) = & \frac{3}{5}(10)-\frac{1}{5}(5) & = 5.\\
v(10,5) = & -\frac{1}{5}(10)+\frac{2}{5}(5) & = 0.\\
u(13,14) = & \frac{3}{5}(13)-\frac{1}{5}(14) & = 5.\\
v(13,14) = & -\frac{1}{5}(13)+\frac{2}{5}(14) & = 3.\\
u(5,10) = & \frac{3}{5}(5)-\frac{1}{5}(10) & = 1.\\
v(5,10) = & -\frac{1}{5}(5)+\frac{2}{5}(10) & = 3.
\end{align*}
Thus the regions vertices in terms of \(u\) and \(v\) are (1,0), (5,0), (5,3), and (1,3). Notice that this is a rectangular region which we can use directly in the setup of the integral.
Finally we setup and evaluate the integral
\begin{align*}
\int_0^3 \int_1^5 (10u+15v)5 \; du \; dv = &\\
5 \int_0^3 \int_1^5 (10u+15v) \; du \; dv = &\\
5 \int_0^3 5u^2+15vu |_1^5 dv = &\\
5 \int_0^3 125+75v-5+15v dv = &\\
5 \int_0^3 120+90v dv = &\\
5 \left( 120v+45v^2 |_0^3 \right) = & 3825.
\end{align*}
For any smooth transformation \(x(u,v,w),\) \(y(u,v,w),\) \(z(u,v,w),\) the triple integral over a region \(S\) is given by
\begin{equation*}
\tint_S f(x(u,v,w),y(u,v,w),z(u,v,w))\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right| \: du \: dv \: dw
\end{equation*}
where \(\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right| \) is the determinant of the \(3 \times 3 \) gradient matrix (Jacobian) given below.
\begin{align*}
\left| \begin{array}{rrr}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\
\frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w}
\end{array} \right| = \\
\left| \frac{\partial x}{\partial u}\left[\left(\frac{\partial y}{\partial v}\right)\left(\frac{\partial z}{\partial w}\right)-\left(\frac{\partial z}{\partial v}\right)\left(\frac{\partial y}{\partial w}\right)\right] \right. \\
-\frac{\partial x}{\partial v}\left[\left(\frac{\partial y}{\partial u}\right)\left(\frac{\partial z}{\partial w}\right)-\left(\frac{\partial z}{\partial u}\right)\left(\frac{\partial y}{\partial w}\right)\right]\\
\left. +\frac{\partial x}{\partial w}\left[\left(\frac{\partial y}{\partial u}\right)\left(\frac{\partial z}{\partial v}\right)-\left(\frac{\partial z}{\partial u}\right)\left(\frac{\partial y}{\partial v}\right)\right] \right|
\end{align*}
