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.
On
In $\mathbb{R}^3$, cylindrical coordinates are a way of expressing any point by its distance from the $z$-axis, $r$, and its angle, $\theta$ measured counterclockwise from the $x$-axis. We can always restrict $r$ to being positive and $\theta$ to being between $0$ and $2\pi$, and still be able to designate any point with these coordinates. Try using limits where $r$ is non-negative because that is assumed when you use $r\,dz\,dr\,d\theta$ at the end of your integral.
In this example, try using the limits \begin{align*} 0\leq\theta <2\pi, \quad 0\leq r \leq 1, \quad \text{and} \quad -\sqrt{2-r^2}\leq z \leq \sqrt{2-r^2} \end{align*}
The $r$ and $\theta$ limits represent a circle of radius 1, which is the projection of our solid into the $xy$ plane, and the $z$ limits contain the volume between the top and bottom of the sphere.
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.