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