Determine the mass of the solid, if the density at a point $P(x,y,z)$ is proportional to the distance from the $xy$ plane.

Question

A solid can be described in three-dimensional space by the following system of equations:

$$x^2 + y^2 + z^2 \leq 1$$ $$z \geq \sqrt{x^2 + y^2}$$

Determine the mass of the solid, if the density at a point $P(x,y,z)$ is proportional to the distance from the $xy$ plane.

My attempt

We have

$x^2 + y^2 +z^2 \leq 1$
$z\geq \sqrt{x^2 + y^2}$
$x = r \cos \theta$
$y = r \sin\theta$
$z=z$
Now $r^2 = x^2 + y^2 \implies r = \sqrt{x^2 + y^2}$

$\rho (p(x,y,z)) \propto d$.

Now $p(x,y,z) = \frac{kd}{\rho}$

Then $d= \frac{M}{V}$

Therefore $M = dV$

On the other hand we have $r^2 \leq 1$ and $z \geq r$

Is this reasoning correct? and if so, how can I proceed to find the solution to the problem? Any help is appreciated.

Answer 1

The density $\rho(x,y,z)$ is equal to $k|z|$, for some constant $k$. Then, the mass of your solid, computed using cylindrical coordinates is$$\int_0^{2\pi}\int_0^{1/\sqrt2}\int_r^{\sqrt{1-r^2}}kz\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=k\frac{\sqrt2\pi}3.$$That $\frac1{\sqrt2}$ is due to the fact that $r^2\leqslant z^2\leqslant1-r^2$ and that therefore $r^2\leqslant1-r^2$, which is equivalent to $r\in\left[0,\frac1{\sqrt2}\right]$.