I have this formula to calculate the volume created by revolving a curve in polar coordinates $r(\theta)$ around the y-axis:
$V=\frac{2\pi}{3}\int_{\alpha}^\beta r^{3}\sin\theta\,d\theta$
I checked this formula using a circle equation $r=R$. I take a quarter of a circle between $0<\theta<\pi/2$, revolve it around the y-axis, multiply by 2 and indeed I get the volume of a sphere, like this:
$V=2\frac{2\pi}{3}\int_{0}^{\frac{\pi}{2}} R^{3}\sin\theta\,d\theta=\frac{4\pi}{3}R^{3}$
But I have a problem with the curve $r=\sin\theta$ between $0<\theta<\pi$.
This curve plots a circle with radius $R=1/2$ on top of the x-axis.
So If you revolve the curve around the y-axis, you should get a sphere with volume:
$V=\frac{4\pi}{3}\frac{1}{2^{3}}=\frac{\pi}{6}$
However, when I use the formula above I get a different result. If I take the curve between $0<\theta<\pi/2$ and revolve it around the y-axis, I get:
$V=\frac{2\pi}{3}\int_{0}^{\frac{\pi}{2}} \sin^{4}\theta\,d\theta=
\frac{2\pi}{3}\int_{0}^{\frac{\pi}{2}} \left(\frac{1-\cos2\theta}{2}\right)^{2}\,d\theta=
\frac{\pi}{6}\int_{0}^{\frac{\pi}{2}} (1-2\cos2\theta+\cos^{2}2\theta)\,d\theta=
\frac{\pi^{2}}{8}$
What am I missing here?
Thanks !
Best Answer
Part of your problem is the confusion that arises with the inconsistencies in the literature regarding the notation for spherical coordinates, i.e., some authors use $r,\theta,\phi$ while others use $r,\phi,\theta$. Your equation for $V$ is correct, but the interpretation of $\theta$ is not. If you derive the equation for volume of revolution in spherical coordinates you can find that
$$V=\frac{2\pi}{3}\int r^{3}\sin\varphi\,d\varphi$$
where $\varphi$ is measured from the vertical axis. It would then be apparent that you would have $r=\sin(\pi/2-\varphi)=\cos\varphi$, then
$$V=\frac{2\pi}{3}\int_0^{\pi/2} \cos^{3}\varphi~\sin\varphi\,d\varphi=\frac{\pi}{6}$$
as expected.
NOTE ADDED FOR CLARIFICATION
If you derive the equation for volume of revolution in polar coordinates as shown here you obtain
$$ V_{y-\text{axis}}=\frac{2\pi}{3}\int r^3 \cos \theta~d\theta\\ V_{x-\text{axis}}=\frac{2\pi}{3}\int r^3 \sin \theta~d\theta $$
which will lead to the same result. In fact, you can now see that you derived the volume for rotation about the $x$-axis.