Surface and volumes of solids of revolution

Question

The lemniscate $\ r^2 =\ a^2 \cos (2\theta) $ revolves about a tangent at the pole. What will be the volume generated?

I know the answer will be $ \frac {\pi a^3}{4}$ But my question is how? Any help possible is requested.

Answer 1

We have that the area of a lemniscate is

$$A=2\left(\dfrac{1}{2}\int r^2d\theta\right)=a^2\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\cos(2\theta)d {\theta}=a^2$$

the centroid of an half is located at $(x_C,0)$ with

$$x_C=\frac{\dfrac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac23 r\,\cos\theta\, r^2d\theta}{a^2/2}=\frac23a\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos\theta\ \cos^\frac32(2\theta) \,d\theta=\frac23a\frac{3\pi}{8\sqrt 2}=\frac{\sqrt 2\pi}{8}a$$

integral evaluation

then by Pappus theorem

$$V=2\pi R\cdot A=2\pi \left(x_C\frac{\sqrt2}2\right)\cdot a^2=2\pi\left(\frac{\sqrt 2\pi}{8}a\frac{\sqrt 2}2\right)a^2=\frac{\pi^2a^3}{4}$$