I was just wondering if someone could please help me check my work/answer for this problem:
Let $W$ be the ice cream cone region bounded above by the hemisphere $z=\sqrt{2-x^2-y^2}$ and below by the cone $z=\sqrt{x^2+y^2}$ and with mass density $\delta(x,y,z)=z$
a) Find the mass of $W$.
b) Find the $z$-coordinate of the center of mass of $W$.
Work for part a): \begin{align} M &= \iiint_{D}\delta \, dV \\ &= \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{\sqrt{2}} z r\, dz\,dr\,d\theta \\ &= \int_{0}^{2\pi} \int_{0}^{1} r\,dr\,d\theta \\ &= \int_{0}^{2\pi} \frac{1}{2} d\theta \\ &= \pi \end{align}
Work for part b): \begin{align} M_{xy} &= \iiint_{d} z\delta\, dV \\ &= \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{\sqrt{2}} z^2 r \,dz\,dr\,d\theta \\ &= \int_{0}^{2\pi} \int_{0}^{1} \frac{2\sqrt{2}}{3} r \,dr\,d\theta \\ &= \frac{2\sqrt{2}}{3} \int_{0}^{2\pi} \frac{1}{2} \,d\theta \\ &= \frac{2\sqrt{2}}{3}\pi \end{align}
$$ \bar{z} = \frac{M_{xy}}{M} = \frac{2\sqrt{2}}{3}\pi \cdot \frac{1}{\pi} = \frac{2\sqrt{2}}{3}. $$
No, that is not correct. As you know, $\theta$ can take any value in $[0,2\pi]$. For each $\theta$, $r$ can take any value in $[0,1]$. But the range of the possible values of $z$ is $\left[r,\sqrt{2-r^2}\right]$. So, for instance, the mass is equal to$$\int_0^{2\pi}\int_0^1\int_r^{\sqrt{2-r^2}}zr\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\frac\pi2.$$