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