Let $f(x,y,z)=z$ and $T=\lbrace(x,y,z) \in \mathbb{R^3} : y\geq0, z\geq0, x^2+y^2+z^2\leq 2, x^2+y^2\leq1\rbrace$
Find $\iiint_T f(x,y,z) dV$
I'm having a few problems with this integral, here's what I've done:
I found that $0\leq z \leq \sqrt{2-x^2-y^2}$.
And then I'm having problems with the $x$-$y$ bounds:
I get two equations: $x^2+y^2\leq2$ and $x^2+y^2\leq1$, and as $y\geq0$ I get this:
I then, convert to polar coordinates, and I get $r\leq \sqrt2$ and $r\leq1$, but with the given $T$, I thought ? that I had to integrate from $r=0$ to $r=\sqrt2$:
$\int_{0}^{\pi}\int_{0}^{\sqrt2}\int_{0}^{\sqrt{2-x^2-y^2}} z$ $dz r dr d\theta = \int_{0}^{\pi}\int_{0}^{\sqrt2}\frac{z^2}{2}\Big|_0^\sqrt{2-x^2-y^2}$ $rdrd\theta=\int_{0}^{\pi}\int_{0}^{\sqrt2}\frac{2-x^2-y^2}{2}$ $rdrd\theta$
At this point I convert $\frac{2-x^2-y^2}{2}$ to polar coordinates:
$\int_{0}^{\pi}\int_{0}^{\sqrt2}\frac{2-r^2}{2}$ $rdrd\theta = \int_{0}^{\pi}\int_{0}^{\sqrt2}\frac{2r-r^3}{2}$ $drd\theta = \int_{0}^{\pi}\frac{r^2}{2}-\frac{r^4}{4}\Big|_0^\sqrt{2}d\theta=\int_{0}^{\pi}\frac{1}{2} d\theta = \frac{\pi}{2}$ which is the wrong answer...
The correct answer would be:
\begin{align} \iiint_T f(x,y,z) dV = \frac{3\pi}{8} \end{align}
$x^{2}+y^{2} \leq 1$ automatically implies $x^{2}+y^{2} \leq 2$ so igonre this second condition. $\theta$ varies from $0$ to $\pi$ and $r$ from $0$ to $1$. Now see if you get the correct answer.