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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 !

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  • $\begingroup$ Is $\theta$ the polar angle of cylindrical coordinate system with $y$ rotation axis, or is it the angle made by radius vector to $x-$axis as polar angle coordinate reference? $\endgroup$
    – Narasimham
    Jan 22 '18 at 17:15
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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.

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  • $\begingroup$ Thanks a lot !! $\endgroup$
    – Shai Yefet
    Jan 24 '18 at 11:30

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