Compute the volume bounded by the parabolic cylinders $x^2=4-4z, \quad y^2=4-4z$ and the $xy-$plane.

Question

Compute the volume bounded by the parabolic cylinders $x^2=4-4z, \quad > y^2=4-4z$ and the $xy-$plane.

Say the $xy-$plane is the ground, the two cylindrcal paraboloids then make upp the roof, where the base is simply a square. Looking at this from above and onto the $xy-$ plane, the following image should emerge:

Hence the volume of this region should easily be computed by

$$V=\iiint_K1 \ dxdydz = 8\iint_D\left(\int_0^{1-x^2/4}1\right)\ dxdy=8\iint_D1-\frac{x^2}{4} \ dydx=$$

$$8\int_{0}^{1}\left(1-\frac{x^2}{4}\right)\int_0^x1 \ dydx= 8\int_0^1x-\frac{x^3}{4} \ dx =\frac{7}{2}.$$

Correct answer: $V=8.$ Why is this?

Answer 1

Note that in the integral $x$ varies between $0$ and $2$, thus

$$8\int_0^2 x-\frac{x^3}{4} \ dx = \left[\frac{x^2}2-\frac{x^4}{16}\right]_0^2=8(2-1)=8$$