Volume of revolution of polar curve

Question

The exercise is to derive the formula for the rotation of a polar curve of the form $ r = r ( \theta ) $ about the $ x $ - axis. When I asked my lecturer, he suggested that I consider a series of infinitely thin cylinders with circumference $ 2 \pi y $, where $ y = r \sin \theta $ denotes the vertical distance from the $ x $ axis and an area given by $ \text{d} A = r \; \text{d} r \; \text{d} \theta $ to obtain $ V = \int r^3 \sin \theta \; \text{d} \theta $, which ended up giving the correct numerical answer. I instead tried converting to Cartesian as below which instead resulted in the following $$ \begin{aligned} V &= \pi \int y^2 \; \text{d} x \\ x &= r \cos \theta \\ \text{d} x &= - r \sin \theta \; \text{d} \theta \\ y^2 &= r^2 \sin ^2 \theta \\ V &= \pi \int r^3 \sin ^3 \theta \; \text{d} \theta \\ \end{aligned} $$. Why does my method not give the correct formula?

Answer 1

Your equation is not quite correct. I approached it this way...

$$V=2\pi\int\!\!\!\int y~dy~dx=2\pi\int\!\!\!\int r\sin\theta\cdot r~dr~d\theta=\frac{2\pi}{3}\int r^3\sin\theta~d\theta$$

You can see how this would work for a sphere by integrating over $\theta\in[0,\pi]$ to get $V=\frac{4\pi r^3}{3}$.