Let $f(x,y,z)=z$ and
$$D=\{(x,y,z)\in\mathbb{R}^3:\ y\geq0,\ z\geq0,\ x^2+y^2+z^2\leq2,\ x^2+y^2\leq1\}.$$
Then
$$\iiint_Df(x,y,z)dxdydz$$ is (hint: use cylindrical coordinates)(A) $\frac{3\pi}{2}\qquad$ (B) $\frac{\pi}{2}\qquad$ (C) $\frac{3\pi}{8}\qquad$ (D) $\frac{5\pi}{6}\qquad$ (E) $\frac{3\pi}{4}$
I need to calculate that integral. We are suggested to use cylindrical coordinates. After a lot of thought I know that the integration domain is the intersection of a radius $\sqrt2$ sphere centered on $(0,0,0)$ and a half cylinder of radius $1$, also centered on $(0,0,0)$.
I noticed that the radius of my cylinder coordinates started to change after $z=1$, so I divided the integral in those two:
$\int_{0}^{1}\int_{0}^{\pi}\int_{0}^{1}z\rho d\rho d\theta dz$, which is the half cylinder part of the domain, and
$\int_{1}^{\sqrt2}\int_{0}^{\pi}\int_{0}^{\sqrt{2-z^2}}z\rho d\rho d\theta dz$, wich is the curved top.
I see it this way:
But after computing everything, it gives me a wrong result. It's my reasoning flawed or I'm computing something wrong? Thank you very much.
Best Answer
An idea for you: in cylindric coordinates, you'll have
$$0\le \theta\le\pi\;\; \text {(just the part for}\;y\ge0)\;,\;\;0\le r\le 1\;\;\text{(within the cylinder)}\,,\;\;0\le \sqrt{2-r^2}$$
the last part being from "the floor" $\;z=0\;$ to the top $\;z=\sqrt{2-x^2-y^2}=\sqrt{2-r^2}\;$ , so your integral becomes
$$\int_0^\pi\int_0^1\int_0^{\sqrt{2-r^2}}\,r\,z\,dz\,dr\,dz$$
with $\;r=\;$ Jacobian. The answer is (c)