Use cylindrical coordinates to find the volume of the solid in the first octant inside of the sphere x^2 + y^2 + z^2 = 4 and inside the cylinder x^2 + y^2 = 2x
Note: set up the integral only, do not evaluate.
The region they have in common is the cross sectional area of the cylinder $\displaystyle (x-1)^2+y^2=1$, thus, we see that $\displaystyle 0\leq x\leq 2$, $\displaystyle -\sqrt{1-(x-1)^2}\leq y\leq \sqrt{1-(x-1)^2}$, and the height of the solid would be defined by $\displaystyle -\sqrt{4-x^2-y^2}\leq z\leq \sqrt{4-x^2-y^2}$
In cylindrical coordinates, we can see that $\displaystyle 1\leq r\leq 2$, $\displaystyle 0\leq\vartheta\leq2\pi$, and $\displaystyle -\sqrt{4-r^2}\leq z\leq \sqrt{4-r^2}$
After a little more work, we see that the volume could be represented as $\displaystyle 2\int_0^{2\pi}\int_1^2 \int_0^{\sqrt{4-r^2}}r\,dz\,dr\,d\vartheta$
Does this make sense?
[I'm pretty sure my limits on $\displaystyle r$ are correct...]
--Chris