# 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.

• do you mean $z \leq c$? – user66081 Jan 28 '14 at 0:11

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.

• Beautifully written explanation of the solution to these types of problems. Like, best source I could find on Google, good. Thank you so much for taking the time to write this. – Defacto Oct 19 '17 at 22:26
• @Someguy Thank you! Your comment made my day! – Mark Fantini Oct 20 '17 at 1:14

Here is a hint, for what it's worth.

The mass in this case is the integral over the solid of the constant function $B$, the center of mass is the integral of the vector-valued function $(x,y,z)$.

You can parameterize the solid, say $V$, in Cartesian coordinates, $$V = \{ (x, y, z) : 0 \leq z \leq c, 0 \leq x^2 + y^2 \leq \frac12 z \}$$ or in polar coordinates $$V = \{ (x, y, z) = (r \cos \phi, r \sin \phi, z) : 0 \leq z \leq c, 0 \leq r^2 \leq \frac12 z, 0 \leq \phi \leq 2\pi \}.$$

However, by symmetry, it is clear that the $x$ and $y$ components of the center of mass will be zero. Hence, the $z$ component of the center of mass will divide the solid into two equal-weight parts above $z$ and below $z$.

You may alternatively consider $V$ as a body of revolution and apply some specialized formula.