Find the volume of $G$ which bounded $z=-\pi,z=\pi$ and the side bounds are $x^2 + y^2 - \cos^{2} z = 1$.

Question

Find the volume of $G$ which bounded $z=-\pi,z=\pi$ and the side bounds are $x^2 + y^2 - \cos^{2} z = 1$.

I have to establish the limits of $x,y,z$, the limits of $z$ are : $-\pi\leq z \leq \pi$ , the limits of $x$ are : $-\sqrt{1-cos^2z-y^2}\leq x \leq \sqrt{1-cos^2z-y^2}$.

Then calculate the triple integral $\int_{-\pi}^{\pi}dz \int_{?}^{?}dy \int_{-\sqrt{1-cos^2z-y^2}}^{\sqrt{1-cos^2z-y^2}} dx$

How am I supposed to find the limits of $y$ ?

Thanks !

Answer 1

This exercise can be solved using a single integral from $z=-\pi$ to $z=\pi$.

A cross-section of the solid parallel to the $xy$ plane intersects the solid in a circle of radius $r=\sqrt{1+\cos^2z}$ and area $A(z)=\pi(1+\cos^2z)$.

Therefore the volume of the solid is given by

$$ V=\pi\int_{-\pi}^\pi1+\cos^2z\,dz $$