Question

Find the volume of the solid in the first octant bounded by the parabolic cylinder z = 4 − x2 and the plane y = 2.

Answer 1

The volume is given by the triple integral,

$\displaystyle\int_0^2\int_0^2\int_0^{4-x^2}\mathrm dz\,\mathrm dx\,\mathrm dy$

Easy enough to compute. Since the integrand is independent of $y$, you can compute that integral right away to get

$2\displaystyle\int_0^2\int_0^{4-x^2}\mathrm dz\,\mathrm dx$

Next, with respect to $z$,

$2\displaystyle\int_0^2z\bigg|_{z=0}^{z=4-x^2}\,\mathrm dx=\int_0^2(4-x^2)\,\mathrm dx$

which results in a volume of

$2\displaystyle\left(4x-\frac13x^3\right)\bigg|_{x=0}^{x=2}=2\left(8-\frac83\right)=\dfrac{32}3$