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{\bf Question} Sketch the region defined by the inequalities
$x^2+y^2\leq z$, $0\leq z\leq 2$. If the region is occupied by a
solid whose density at the point $(x,y,z)$ is $(3-z)$, calculate
its total mass by means of a triple integral. (HINT: Transform to
cylindrical co-ordinates.)



{\bf Answer}

For each z, $x^2 + y^2 \leq z$ is a disc of radius $\sqrt z$.  If
we let z vary as $0 \leq z \leq 2$ we obtain the region:

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Volume of region = $\int \!\!\! \int \!\!\! \int (3 - z)d(x,y,z)$

We use the cylindrical coordinates $(\rho, \phi\ z)$ where $x =
\rho \cos \phi$ and $y = \rho \sin \phi$.  Since $x^2 + y^2 =
\rho^2(\cos^2 \phi + \sin^2 \phi) = \rho^2$, the region is defined
by the inequalities: $\rho^2 \leq z \Rightarrow 0 \leq \rho \leq
\sqrt z$, \hspace{.1in}   $0 \leq z \leq 2$ with any $\phi$ so
that $0 \leq \phi \leq 2\pi$,  Using $d(x,y,z) = \rho \, d\phi \,
d\rho \, dz$

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\begin{eqnarray*}
{\rm Volume\ of\ region\ } & = & \int_{z = 0}^{z = 2} \!
\int_{\rho = 0}^{\rho = \sqrt z} \! \int_{\phi = 0}^{\phi = 2\pi}
(3 - z)\rho \, d\phi \, d\rho \, dz \\ & = & \int_{z = 0}^{z = 2}
\! \int_{\rho = 0}^{\rho = \sqrt z} \left[(3 - z)\rho
\phi\right]_{\phi = 0}^{\phi = 2\pi} \, d\rho \, dz \\ & = & 2\pi
\int_{z = 0}^{z = 2} \! \int_{\rho = 0}^{\rho = \sqrt z} (3 -
z)\rho \, d\rho \, dz
\\  & = & 2\pi \int_{z = 0}^{z = 2} \left[\frac{1}{2}(3 - z)\rho^2
\right]_{\rho = 0}^{\rho = \sqrt z}  dz \\ & = & \pi \int_{0}^{2}
(3 - z) z dz
\\ & = & \pi \int_0^2 3z - z^2 dz \\ & = & \pi\left[\frac{3}{2} z^2 -
\frac{1}{3} z^3 \right]_0^2 \\ & = & \pi\left[6 -
\frac{8}{3}\right]
\\ & = & \frac{10\pi}{3}
\end{eqnarray*}


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