I am trying to solve the following problem*
A solid can be described in three-dimensional space by the following system of inequalities:** $$\begin{cases} x^2+y^2+z^2\leq1\\ z\geq\sqrt{x^2+y^2} \end{cases}$$ Determine the mass of the solid, provided its density at a point $P(x,y,z)$ is proportional to the distance from the $XY$ plane.
I came out with the following triple integral
$$k\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{\sqrt{1-r^2}} rz dz dr dθ. $$
but the problem is that I got zero where I calculed the integral because \begin{align*} M &= k \int_{0}^{2\pi} \int_{0}^{1} \left[ \frac{1}{2}rz^2 \right]_{r}^{\sqrt{1 - r^2}} \, dr \, d\theta \\ &= k \int_{0}^{2\pi} \int_{0}^{1} \left( \frac{1}{2}r(1 - r^2) - \frac{1}{2}r^3 \right) \, dr \, d\theta \\ &= k \int_{0}^{2\pi} \left[ \frac{1}{4}r^2 - \frac{1}{4}r^4 \right]_{0}^{1} \, d\theta \\ &= k \int_{0}^{2\pi} \left( \frac{1}{4} - \frac{1}{4} \right) \, d\theta \\ &= k \int_{0}^{2\pi} 0 \, d\theta \\ &= 0. \end{align*}
Did I do something wrong and where? Thanks!
Note that $$\begin{cases} x^2+y^2+z^2\leq1\\ z\geq\sqrt{x^2+y^2} \end{cases}\implies r^2\leq z^2\leq 1-r^2\implies r^2\leq \frac{1}{2}.$$ Therefore your integral should be $$\begin{align*} M &=k\int_{0}^{2\pi}\int_{0}^{\color{blue}{1/{\sqrt{2}}}}\int_{r}^{\sqrt{1-r^2}} rz dz dr d\theta\\ &= k \int_{0}^{2\pi} \int_{0}^{\color{blue}{1/{\sqrt{2}}}} \left[ \frac{1}{2}rz^2 \right]_{r}^{\sqrt{1 - r^2}} \, dr \, d\theta \\ &= k \int_{0}^{2\pi} \int_{0}^{\color{blue}{1/{\sqrt{2}}}} \left( \frac{1}{2}r(1 - r^2) - \frac{1}{2}r^3 \right) \, dr \, d\theta \\ &= k \int_{0}^{2\pi} \left[ \frac{1}{4}r^2 - \frac{1}{4}r^4 \right]_{0}^{\color{blue}{1/{\sqrt{2}}}} \, d\theta \\ &= \frac{k}{4} \int_{0}^{2\pi} \left( \color{blue}{\frac{1}{2} - \frac{1}{4}} \right) \, d\theta =\color{blue}{\frac{\pi k}{8}}. \end{align*}$$