volume of revolution with a circle

Hi,

Suppose I have to find the volume of revolution of a unit circle (centre 0,0 radius r) for the first quadrant when it's turned about the x axis.

As $x^2 + y^2=r^2 \rightarrow y^2=r^2 - x^2$ and I need to set the intergrand up in terms of $y^2$ I have:

$\pi \int_0^{r} r^2 - x^2 dx$, but then when I integrate and sub in r and 0 respectively for x, I get 0. Have I arrived at the
wrong intergrand?

When I sub in r or 0 for x I get $\pi (\frac{r^3}{3}-\frac{r^3}{3}) - \pi(\frac{r^3}{3} - \frac{0^3}{3} )$ So not 0 but $-\pi \frac{r^3}{3}$

Am I being totally stupid?!

I would not say totally

In the integrand $r^2$ is a constant so the integral is $r^2x$ - you have a $\frac{1}{3}$ there suggesting you turned this into $\frac{r^3}{3}$

...but of course! Yes, totally stupid.


Thanks Sir!!!!

Madam