I'm having trouble with a double integral problem in cylindrical coordinates. I'm sure the answer is staring me in the face, but I'm missing something. In the multivariable version of the Community Calculus textbook, exercise 15.2.2 states asks us to find the volume inside both $r = 1$ and $r^2 + z^2 = 4$.
I solved the double integral as follows:
$$\int_0^{2\pi} \int_0^1 \sqrt{4-r^2}\ r\ dr\ d\theta$$
$$= \int_0^{2\pi} -\frac{1}{3}(4 - r^2)^\frac{3}{2}|_0^1\ d\theta$$
$$= \int_0^{2\pi} -\sqrt{3} + \frac{8}{3}\ d\theta $$
Finally, I solved the outer integral.
$$ = -\sqrt{3}\theta + \frac{8}{3}\theta\ |_0^{2\pi} $$ $$ = \frac{16\pi}{3} - 2\pi\sqrt{3}$$
Unfortunately, my answer is half of what is listed in the solution section. The correct solution is: $$\frac{32\pi}{3} - 4\pi\sqrt{3} $$
Realizing that when I solved the original equation for $z$, the answer was: $$z = \pm\sqrt{4 - r^2}$$
Is that where I went wrong?
Yes, the integral you've written down accounts only for the top half of the solid, which is bounded above and below by the sphere $r^2 + z^2 = 4$, that is by the graphs of the functions $(r, \theta) \mapsto \pm\sqrt{4-r^2}$. The height of an infinitesimal parallelepiped is thus $$\sqrt{4 - r^2} - \left(-\sqrt{4 - r^2}\right) = 2\sqrt{4 - r^2},$$ so the parallelepiped itself has volume $$2\sqrt{4 - r^2} \cdot r \,dr \,d\theta.$$