Question: Find the volume of the intersection of the sphere x^2 + y^2 + z^2 = 2 and the cylinder x^2 + y^2 = 1.
So I get my triple intgral set up as (I am using cylindrical coordinates):
$$\iiint_ \mu(r)\,dz\,dr\,d\theta$$
where limits are:
0<= $\theta$ <=$\pi/2$ , -1<= r <=1 , $-\sqrt{2-r^2}$ <= z <=$\sqrt{2-r^2}$
I end up getting 0 as volume after calulating. Any idea's? I think i am setting up my integral wrong, but I am not sure what I am doing wrong.
Best Answer
Using cylindrical coordinates and symmetry we get the following:
Go up the $z$-axis, hitting $z=0$ first and then exiting at $z=\sqrt{2-r^2}$. So, the projection is two circles, one with $r=1$ and the other $r = 0$. Our volume then becomes $$2\int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{\sqrt{2-r^2}} rdzdrd\theta $$ $$\Rightarrow 4\pi\int_{0}^{1} r\sqrt{2-r^2}dr$$ Using the substitution $u = 2-r^2$ and solving, we get the answer as $\frac{4\pi}{3}(2\sqrt{2}-1)$. Hope it helps.