Finding Mass of Object Given Density

Question

I need to find the mass of an object that lies above the disk $x^2 +y^2 \le 1$ in the $x$-$y$ plane and below the sphere $x^2 + y^2 + z^2 = 4$, if its density is $\rho(x, y, z)=2z$.

I know that the mass will be $\iiint_R 2z$ $dV$, and I just need to determine the region $R$ which bounds the object. If I were to use spherical coordinates, I'd have $(r, \theta, \phi)$ where $0 \le r \le 1$ (since the radial distance is restricted by the disk), $0 \le \theta \le 2\pi$ (since we can complete a full revolution just fine), however I am unsure how to determine the upper limit of $\phi$.

Am I going in the right direction using spherical coordinates? And how would I find the upper limit of $\phi$? Thanks.

Answer 1

It's fine to use spherical coordinates, but I think that it is more natural to use cylindrical ones:\begin{align}\iiint_V2z\,\mathrm dx\,\mathrm dy\,\mathrm dz&=\int_0^{2\pi}\int_0^1\int_0^{\sqrt{4-r^2}}2zr\,\mathrm dz\,\mathrm dr\,\mathrm d\theta\\&=2\pi\int_0^1r(4-r^2)\,\mathrm dr.\end{align}Can you take it from here?

Answer 2

Try to do it using cylindrical coordinates instead.

You can find the limits of $z$ by solving the sphere and a cylinder (extended into z - axis from disc) of radius $r$.

$$(x^2+y^2) + z^2 = 4$$

$$r+z^2 = 4$$ $$z = \sqrt{4-r^2}$$

$$\iiint_V \rho(x,y,z)\,dx\,dy\,dz = \int_0^1\int_0^{2\pi} \int_0^{\sqrt{4-r^2}} 2z\,dz\,(rd\theta)\,dr $$

$$M = 2\pi\int_0^1 r(4-r^2)\,dr = 2\pi\left( 2-\frac{1}{4} \right) = \frac{7\pi}{2} $$