Question

Evaluate the given integral by making an appropriate change of variables. 10 x ? 5y 8x ? y da, r where r is the parallelogram enclosed by the lines x ? 5y = 0, x ? 5y = 2, 8x ? y = 5, and 8x ? y = 10

Answer 1

Take

$\begin{cases}u=x-5y\\v=8x-y\end{cases}$

so that you have

$\begin{cases}\mathbf x(u,v)=\dfrac{-u+5v}{39}\\\\\mathbf y(u,v)=\dfrac{-8u+v}{39}\end{cases}$

which gives a Jacobian determinant of

$|\det J|=\left|\begin{vmatrix}\mathbf x_u&\mathbf x_v\\\mathbf y_u&\mathbf y_v\end{vmatrix}\right|=\dfrac1{32}$

So upon transforming the coordinates to the u-v plane, you have (and I'm guessing on what the integrand actually is)

$\displaystyle\iint_R(10x-5y)(8x-y)\,\mathrm dA=\frac1{32}\int_{u=0}^{u=2}\int_{v=5}^{v=10}\frac5{13}(2u+3v)v\,\mathrm dv\,\mathrm du$
$=\displaystyle\frac5{416}\int_{u=0}^{u=2}\int_{v=5}^{v=10}(2uv+3v^2)\,\mathrm dv\,\mathrm du=\dfrac{2375}{104}$