Question

Use the given transformation to evaluate the integral. 9 x − 3y 2x − y dA, R where R is the parallelogram enclosed by the lines x − 3y = 0, x − 3y = 10, 2x − y = 9, and 2x − y = 10; u = x − 3y, v = 2x − y

Answer 1

$\displaystyle\iint_R(9x-3y)(2x-y)\,\mathrm dA$

$\begin{cases}u=x-3y\\v=2x-y\end{cases}\implies\begin{cases}x=\frac{3v-u}5\\y=\frac{v-2u}5\end{cases}\implies\mathrm dA=|\det J|\,\mathrm du\,\mathrm dv$

where $J$ is the Jacobian matrix for the transformation,

$J=\begin{bmatrix}u_x&u_y\\v_x&v_y\end{bmatrix}=\begin{bmatrix}-\dfrac15&\dfrac35\\\\-\dfrac25&\dfrac15\end{bmatrix}\implies|\det J|=\dfrac15$

We have

$9x-3y=8x+x-3y=\dfrac85(3v-u)+u=\dfrac{24v-3u}5$

so that the integral is

$\displaystyle\frac1{25}\int_9^{10}\int_0^{10}(24v-3u)v\,\mathrm du\,\mathrm dv=\boxed{\frac{4051}5}$