Calculating the volume of $E,$ where $E$ is bounded by $x^2+y^2+z^2=4$ and $x^2+y^2=2y.$
I've changed this to cylinder coordinate ; $\theta: 0-\pi , r : 0-2\sin(\theta), z : 0-\sqrt{4-r^2}$
then the gorgeous "wolfram alpha" gave the result : $\frac {8\pi}{3} - \frac {32}{9}$
and it's right ans
I wanna know my prob. solving this
$\iint r \sqrt{4-r^2}\ dr\ d\theta$
$=\int 8(\frac {1-\cos^3\theta}{3})\ d\theta$
$=\frac {8\pi}{3}$ .....?! (Wrong)
HINT: It appears that you think that the integral of $-\frac{8}{3} \cos^3 \theta$ from $0$ to $\pi$ is zero. You might want to check that.