Is there a way to solve a volume of revolution question using cylindrical coordinates by using three iterated integrals $dr,dz,d\theta$ ? This is the question:
I can solve the volume when this is rotates about the y-axis ( the volume only for the region bounded by $PQR$), using the volume of revolution method learned in high school, but I am having trouble setting up the iterated integral for when it is converted to a cylindrical coordinate system. I will appreciate the help.
In the first place, since this is a body of revolution, you only need polar coordinates, not cylindrical coordinates. The solution you are seeking is best explained in terms of Pappus's $2^{nd}$ Centroid theorem: the 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., $2πR$. The bottom line is that the volume is given simply by $V=2\pi RA$. And the best way to approach the polar coordinate system is through the complex plane.
The area and centroid are expressed as follows in the complex plane:
$$ A=\frac{1}{2}\int\Im\{z^*\dot z\}d\theta\\ R=\frac{1}{3A}\int z\ \Im\{z^*\dot z\}d\theta $$
Thus, the volume is given by
$$ V=\frac{2\pi}{3}\int z\ \Im\{z^*\dot z\}d\theta $$
Now, allowing that $z=re^{i\theta}$, we find $$ \dot z=(\dot r + ir)e^{i\theta}\\ z^*=re^{-i\theta}\\ z^*\dot z=r(\dot r+ir)\\ \Im\{z^*\dot z\}=r^2\\ $$
There follows,
$$ V=\frac{2\pi}{3}\int r^3e^{i\theta} d\theta $$
And transforming this back from complex to polar coordinates, we get the results for rotation about the $y-$ and $x-$axes as follows:
$$ V_{y-\text{axis}}=\frac{2\pi}{3}\int r^3 \cos \theta~d\theta\\ V_{x-\text{axis}}=\frac{2\pi}{3}\int r^3 \sin \theta~d\theta $$