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{\bf Exam Question

Topic: TripleIntegral}

A sphere $S$ whose equation is $x^2+y^2+z^2=a^2$ has density which
is proportional to the square of the distance from the plane
$z+a=0,$ i.e. the plane which is tangent at the \lq\lq south
pole". Find the total mass of the sphere.

 \vspace{0.5in}

 {\bf Solution}

The density is given by $\rho(x,y,z)=k(a+z)^2.$ The mass is given
by $$M=\int \!\!\!\int \!\!\!\int_S k(a+z^2)\, dV=\int \!\!\!\int
\!\!\!\int_S a^2\, dV+2ka\int \!\!\!\int \!\!\!\int_S z\, dV+k\int
\!\!\!\int \!\!\!\int_S z^2\, dV.$$

Now $\displaystyle \int \!\!\!\int \!\!\!\int_S z\, dV=0$ by
symmetry.

Also $\displaystyle \int \!\!\!\int \!\!\!\int_S a^2\,
dV=a^2.\frac{4}{3}\pi a^3=\frac{4}{3}a^5.$

Finally we have, using spherical polar coordinates,
\begin{eqnarray*}
\int \!\!\!\int \!\!\!\int_S z^2\, dV &amp;=&amp; \int_0^{2\pi}\,
d\phi\int_0^{\pi}\, d\theta\int_0^a
(r\cos\theta)^2.r^2\sin\theta\, dr\\
&amp;=&amp;2\pi\int_0^{\pi}\sin\theta\cos^2\theta\,d\theta\int_0^a r^4\,
dr\\
&amp;=&amp;2\pi\left[\frac{\cos^3\theta}{3}\right]_0^{\pi}.\frac{a^5}{5}=\frac{4}{15}\pi
a^5.
\end{eqnarray*}




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