Finding the Centroid of Solid G?

Question

I saw this problem on one of my assignments and had no idea how to do it, mostly because I missed the section where it was covered. Anyways it states: Find the centroid of solid G defined by the inequalities $\sqrt{x^2+y^2}$ $\le$ z $\le$ $20-x^2-y^2$. Find the coordinates of the centroid of G.

How should I approach this problem and others that could be similar to it. Thanks

Answer 1

Hint: Supposing the density is uniform across the solid, the centroid is simply calculated by

$\bar x = {{\iiint x dV} \over {\iiint dV}}$

$\bar y = {{\iiint y dV} \over {\iiint dV}}$

$\bar z = {{\iiint z dV} \over {\iiint dV}}$

Now you are going to calculate these volume integrals.

P.S If the density is not a constant, i.e. , $d=f(x,y,z)$, you should place it before $dV$ in every integrand. For example,

$\bar x = {{\iiint x f(x,y,z) dV} \over {\iiint f(x,y,z) dV}}$

Answer 2

From Wikipedia (https://en.wikipedia.org/wiki/Center_of_mass): If the mass distribution is continuous with the density $ρ(r)$ within a volume $V$, then the integral of the weighted position coordinates of the points in this volume relative to the center of mass $R$ is zero, that is

$$\int\limits_{V}{\rho \left( \mathbf{r} \right)\left( \mathbf{r-R} \right)dV}\,=0 $$ Clearly, then, for a volume of uniform density

$$\mathbf{R}=\frac{1}{V}\int\limits_{V}{\mathbf{r}dV}=\frac{1}{V}\int\int\int(x\mathbf{i}+y\mathbf{j}+z \mathbf{k})\,dx\,dy\,dz $$

where, of course

$$V=\int\int\int dx\,dy\,dz $$

Answer 3

From symmetry, the centroid $G$ will be on the $z$ axis. That is,

$G = (0, 0, \overline{z} ) $

Now,

$ \overline{z} = \dfrac{1}{V} \displaystyle \iiint z dV $

Since the object is circularly symmetric about the origin, it is best to use cylindrical coordinates. Thus $ x = r \cos \phi $, $ y = r \sin \phi$, $z = z$

Now, we have two functions $z = r$ and $z = 20 - r^2 $, to find the upper bound of $r$ set them equal

$ r = 20 - r^2 \Longrightarrow r^2 + r - 20 = 0 \Longrightarrow (r + 5)(r - 4) = 0 \Longrightarrow r = 4 $

Therefore,

$ V = \displaystyle \int_{\phi = 0 }^{2 \pi} \int_{r = 0 }^4 \int_{z=r }^{20 - r^2} dz r dr d \phi = 2 \pi \displaystyle \int_{r = 0}^4 20r - r^2 - r^3 dr = 2 \pi \bigg[10 r^2 - \dfrac{1}{3} r^3 - \dfrac{1}{4} r^4 \bigg]_0^4 $

So that,

$ V = \dfrac{448 \pi}{3} $

Now,

$\begin{equation} \begin{split} \displaystyle \iiint z dV &= \displaystyle \int_{\phi = 0 }^{2 \pi} \int_{r = 0 }^4 \int_{z=r }^{20 - r^2} z dz r dr d \phi \\ &= 2 \pi \displaystyle \int_{r = 0}^4 \dfrac{1}{2} r ( (20 - r^2)^2 - r^2 ) dr \\&= \pi \bigg[ \dfrac{r^6}{6} - \dfrac{41 r^4}{4} + 200 r^2 \bigg]_0^4 \\ &= \dfrac{3776 \pi}{3}\end{split}\end{equation} $

Hence,

$ \overline{z} = \dfrac{3776}{448} = \dfrac{59}{7} \approx 8.42857 $