# volume of revolution given by $r= \cos (\theta), z= \sin(2\theta)$ about z-axis

The problem which I want to solve is to find:

Volume of revolution obtained by rotating the area bounded by below curves: \begin{align*} r= \cos \theta, \quad z= \sin (2\theta), \quad -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \end{align*} about $$z$$ axis. Intutively I thought that the solid which obatined by above must be like a donut, just without the hole(this may wrong). It is like rotating a slanted disk along vertical axis. My text book says that I have to use divergence theorem to solve the problem, by parametrizing the surface of the revolution. but I have difficulty in parametrization since I have to parametrize the already parametrized object. The given curve is obviously parametrized as \begin{align*} (x = \cos^2 \theta, \quad y= \sin \theta \cos \theta, \quad z= \sin 2\theta). \end{align*} But now, how should I represent the rotated surface? I am stucked here and need some help!

Your confusion comes from the fact that you also used $$\theta$$ for the angle in the $$xy$$ plane. Your surface should be parametrized as $$(x=\cos\theta\cos\phi, y=\cos\theta\sin\phi, z=\sin(2\theta))$$ Here $$\phi$$ varies between $$0$$ and $$2\pi$$. I think you can use cylindrical shells. Take a cylindrical shell around the $$z$$ axis, of radius $$r$$ and thickness $$dr$$. The height is given by $$z=\sin(2\theta)=2\sin\theta\cos\theta=\pm2r\sqrt{1-r^2}$$ so $$h=4r\sqrt{1-r^2}$$ Then the volume is $$V=2\pi\int_0^1dr\ r\cdot 4r\sqrt{1-r^2}$$
• As you told, the fact that I already used $\theta$ gave me a great confusion. I ve also checked the volume by divergence theorem and it worked! Thanks for the great hint! Mar 18, 2020 at 20:26