Question

Find the volume of the solid under the surface z = 5x + 9y 2 and above the region bounded by x = y 2 and x = y 3.

Answer 1

Call the region in the $x$-$y$ plane, bounded by $x=y^2$ and $x=y^3$, $\mathcal D$. Then the volume under the given surface is

$\displaystyle\iint_{\mathcal D}(5x+9y)\,\mathrm dA=\int_{y=0}^{y=1}\int_{x=y^3}^{x=y^2}(5x+9y)\,\mathrm dx\,\mathrm dy=\frac{83}{140}$