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