Question

Evaluate the given integral by making an appropriate change of variables. 7 x − 4y 7x − y dA, R where R is the parallelogram enclosed by the lines x − 4y = 0, x − 4y = 5, 7x − y = 1, and 7x − y = 4

Answer 1

The obvious substitution is

$\begin{cases}u=x-4y\\v=7x-y\end{cases}$

so that the limits of integration in the $u,v$-plane are constant.

We can solve the above system for $x,y$ to get

$\begin{cases}x=-\frac1{27}(u-4v)\\y=-\frac1{27}(7u-v)\end{cases}$

The Jacobian for this change of coordiantes is

$J=\begin{bmatrix}\dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=-\dfrac1{27}\begin{bmatrix}1&-4\\7&-1\end{bmatrix}$

so that the differential becomes

$\mathrm dx\,\mathrm dy=|\mathrm{det}(J)|\,\mathrm du\,\mathrm dv=\dfrac1{27}\,\mathrm du\,\mathrm dv$

Now,

$\displaystyle\iint_R(7x-4y)(7x-y)\,\mathrm dA=\iint_R(6x+(x-4y))(7x-y)\,\mathrm dA$

$=\displaystyle\frac1{27}\int_1^4\int_0^5\left(-\frac6{27}(u-4v)+u\right)v\,\mathrm du\,\mathrm dv$

$=\displaystyle\frac1{243}\int_1^4\int_0^5(7u+8v)v\,\mathrm du\,\mathrm dv$

$=\displaystyle\frac1{243}\int_1^4\left(\frac{175}2v+40v^2\right)\,\mathrm dv=\boxed{\frac{665}{108}}$