Question

Find the volume of the solid under the plane 3x+2y−z=0 and above the region enclosed by the parabolas y=x^2 and x=y^2.

Answer 1

The region in the x-y plane is the set of points

$R=\{(x,y)\mid0\le x\le 1,x^2\le y\le\sqrt x\}$

The height of the solid falls between the x-y plane (for which z = 0) and the equation of the plane, z = 3x + 2y.

So the volume is

$\displaystyle\int_0^1\int_{x^2}^{\sqrt x}\int_0^{3x+2y}\mathrm dz\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^1\int_{x^2}^{\sqrt x}(3x+2y)\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^1(3x^{3/2}+x-3x^3-x^4)\,\mathrm dx$

$=\displaystyle\frac65+\frac12-\frac34-\frac15=\boxed{\dfrac34}$