Use a change of variables to find the volume of the solid region lying below the surface $z = f(x, y)$ and above the plane region $R$.

Question

Use a change of variables to find the volume of the solid region lying below the surface $z = f(x, y)$ and above the plane region $R$.
$$f(x, y) = (7x + 2y)^2 \sqrt{2y-x}.$$

$R$: region bounded by the parallelogram with vertices $(0, 0), (-1,7/2), (2,5), (3,3/2)$

Answer 1

The parallelogram is bounded by the lines:
$7x +2y = 0\\ 2y-x = 0\\ 2y-x = 4\\ 7x+2y = 24$

Let $u = x-2y, v = 7x+2y$

$du\ dv = \left|\begin{matrix}7&2\\-1&2 \end{matrix}\right| \ dy\ dx= 16 \ dy\ dx$

$\int_0^{24}\int_0^4 \frac {u^2v^\frac 12}{16} \ dv\ du$