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.

1 answer:

8 0

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

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

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