Center of Mass, Multivariable Calculus I have a solid with the bounds $z=2x^2+2y^2$ where $z=c$ and this solid has a uniform density of B. I need to find the mass and the center of mass of this solid. I know how to find a normal center of mass, but I do not know how to set up an integral for this problem, but I think it involves change of coordinates (Also, assume c>0). Thanks. 

The mass of the solid is defined as
$$M = \iiint\limits_{\mathcal{B}} \rho \, dV,$$
that is, the integral of body density at each point over the volume of the body. In this case we have $\rho \equiv B$ which is constant, therefore the mass will be a multiple of the body's volume:
$$M = \iiint\limits_{\mathcal{B}} \rho \, dV = B \iiint\limits_{\mathcal{B}} \, dV.$$
This is a paraboloid and its volume can be found using the cylindrical coordinate substition. We find $z = 2(x^2+y^2) = 2r^2$ and the limits for $z$ will be $2r^2 \leq z \leq \sqrt{c/2}$. This was found equating $z=2r^2 = c$, therefore $r = \sqrt{c/2}$.
$$
\begin{align}
\iiint\limits_{\mathcal{B}} \, dV & = \int_0^{2 \pi} \hspace{-5pt} \int_0^{\sqrt{c/2}} \hspace{-5pt} \int_{2r^2}^c r \, dz \, dr \, d \theta \\
 & = 2 \pi \int_0^{\sqrt{c/2}} cr - 2r^3 \, dr \\
 & = 2 \pi \left( \frac{cr^2}{2} - \frac{r^4}{2} \right) \Bigg\vert_0^{\sqrt{c/2}} \\
 & = \pi \left( c \cdot \frac{c}{2} - \frac{c^2}{4} \right) \\
 & = \frac{c^2 \pi}{4}.
\end{align}
$$
Hence $M = (Bc^2 \pi)/4$.
Coordinates for center of mass are defined as
$$
\begin{align}
\overline{x} & = \frac{1}{M} \iiint x \rho \, dV \\
\overline{y} & = \frac{1}{M} \iiint y \rho \, dV \\
\overline{z} & = \frac{1}{M} \iiint z \rho \, dV.
\end{align}
$$
In physicist notation I've seen this written as
$$\mathbf{R} = \frac{1}{M} \iiint \rho \, \mathbf{r} \, dV.$$
It is interesting to note the following: the paraboloid is symmetric around $z$ axis. This means that the center of mass must be in the $z$ axis, for the $\overline{x}$ and $\overline{y}$ will cancel (if you don't believe this, write out the integral explicitly: you will have to integrate $\cos \theta$ and $\sin \theta$ over $[0,2 \pi]$, which is zero).
Therefore it is left for us to compute the $z$ coordinate. Leaving out the density out for a second (since it is uniform), we have
$$
\begin{align}
\iiint\limits_{\mathcal{B}} z \, dV & = \int_0^{2\pi} \hspace{-5pt} \int_0^{\sqrt{c/2}} \hspace{-5pt} \int_{2r^2}^c z r \, dz \, dr \, d \theta \\
 & = 2 \pi \int_0^{\sqrt{c/2}} \frac{r}{2} \left( c^2 - 4r^4 \right) \, dr \\
 & = \pi \int_0^{\sqrt{c/2}} c^2 r - 4r^5 \, dr \\
 & = \pi \left( \frac{(cr)^2}{2} - \frac{2r^6}{3} \right) \Bigg\vert_0^{\sqrt{c/2}} \\
 & = \pi \left( \frac{c^2}{2} \cdot \frac{c}{2} - \frac{2}{3} \cdot \frac{c^3}{8} \right) \\
 & = \pi \left( \frac{c^3}{4} - \frac{c^3}{12} \right) \\
 & = \pi \left( \frac{c^3}{6} \right) \\
 & = \frac{c^3 \pi}{6}.
\end{align}
$$
Finally
$$\overline{z} = \frac{B c^3 \pi}{6} \cdot \frac{4}{Bc^2 \pi} = \frac{2c}{3}.$$
Therefore
$$M = \frac{Bc^2 \pi}{4} \text{ and } \mathbf{R} = (\overline{x}, \overline{y}, \overline{z}) = \left( 0, 0, \frac{2c}{3} \right).$$
Hope this helps and best wishes.
