Volume of revolution of polar curve 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?
 A: 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 $V=\frac{4\pi r^3}{3}$.
A: What about deriving this formula without using double integrals. One now has to show that a typical conical shell generated by rotating about the x-axis an infinitesimal arc of the curve  r=r(θ) (corresponding to infinitesimal dθ)  has infinitesimal volume
dV=1/3 (2πrsinθ)⋅r ⋅rdθ
A: A curve in polar coordinates is defined by $r=f(\theta)$, where $f$ is a function.
If $y=f(\theta)\cos(\theta)\ge 0$ over the interval $[\alpha, \beta]$, let $R$ be the region bounded by $r=f(\theta), \theta=0, x=a=f(\alpha)\cos(\alpha)$ and $x=b=f(\beta)\cos(\beta)$. The volume of the solid obtained by revolving $R$ about the $x$-axis is given by 
\begin{eqnarray}
V_{x}&=&\int_a^by^2dx\\
&=&\int_{\alpha}^{\beta}[f(\theta)\sin(\theta)]^2x’(\theta)d\theta\\
&=&\int_{\alpha}^{\beta}[f(\theta)\sin(\theta)]^2[f’(\theta)\cos(\theta)-f(\theta)\sin(\theta)]d\theta\\
\end{eqnarray}
