# Centroids and volume

I'm not quite sure what i'm doing wrong here..

Find the centroid $(\bar{x},\bar{y})$ of the plane region defined by:

$$0 \leq y \leq \frac{9-x^2}{9}$$

Then use Pappu's theorem to find the volume $V$ of the solid obtained by rotating the region about the $x$-axis.

What i've done:

$$A=\int_0^3 \frac{9-x^2}{9}$$

$$Mx.0=\int_0^3 x\frac{9-x^2}{9}$$

$$My.0=\frac{1}{2} \int_0^3 \left(\frac{9-x^2}{9}\right)$$

Solving these integrals $\bar{x}$ and $\bar{y}$ should be easy to find.. what i'm not quite sure about is the volume, consindering it's rotation about the $x$-axis i've used $\bar{y}$ as $r$ in the formula $V=2\pi r A$.

What's wrong?

You ask, "What's wrong?" The answer is that you are pulling equations out of the air from memory that may or may not be correct. It may help to take a step back and look at the basic definitions. The area and centroid are given by

$$A=\int\!\!\!\int dy~dx=\int y(x)~dx\\ R_x=\frac{\int\!\!\!\int x~dy~dx}{\int\!\!\!\int dy~dx}=\frac{1}{A}\int x~y(x)~dx\\ R_y=\frac{\int\!\!\!\int y~dy~dx}{\int\!\!\!\int dy~dx}=\frac{1}{2A}\int y^2(x)~dx\\$$

And finally, Pappus's $2^{nd}$ Centroid Theorem states tht he volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $V=2πRA$. Therefore, for rotation about the $x$-axis, we can say that

$$V=\pi\int y^2(x)~dx$$

I'm going to assume here that you intend the full width of curve for $y>0$. Thus, in your case we have

\begin{align}V &=\pi\int_{-3}^3 \left( \frac{9-x^2}{9}\right)^2~dx\\ &=\frac{16\pi}{5}\approx10.053 \end{align}

Firstly, check your limits of integration again. That region is bounded below by the x-axis, and above by the curve. Therefore, the area should be:

\begin{align} Let \quad f(x) &= \frac{9-x^2}{9} \\ \\ A &= \int_{-3}^3 f(x)dx \\ \bar x &=\frac{1}{A} \int_{-3}^3xf(x)dx = 0 \quad \text{Due to symmetry} \\ \\ \bar y &= \frac{1}{A}\int_{-3}^{3} f(x)^2dx \end{align}

The volume of revolution is given by:

$$V=\int_{-3}^3 \pi f(x)^2dx$$

Since the volume of a single disk of revolution at $x$ has a "radius" of $f(x)$, the differential volume element is $dV = \pi f(x)^2 dx$. Then, it's just a matter of adding up all of the small disks from $x=-3$ to $x=3$.

• Can't I use Pappu's theorem? Where bar-r is equal to bar-y?
– Migr
Commented Nov 8, 2013 at 21:41
• I now noticed that if I mutiply the volume by 2 I get the correct answer. But why do I have to multiply it by 2?
– Migr
Commented Nov 9, 2013 at 10:44
• Anyone able to help? =)
– Migr
Commented Nov 10, 2013 at 12:51
• Shouldn't it be $$\bar y = \frac{1}{2A}\int_{-3}^{3} f(x)^2dx$$
– Hrhm
Commented Sep 15, 2016 at 18:43