How to find the surface area of revolution of a polar curve

The formulas we’ll use to find the surface area of revolution of a polar curve

We can find the surface area of the object created when we rotate a polar curve around either the $x$ -axis or the $y$ -axis using the formulas

We can solve for $x$ and $y$ as needed using the conversion formulas

$x=r\cos{\theta}$

$y=r\sin{\theta}$

Example

Find the surface area of revolution of the polar curve over the interval $0\le\theta\le\pi$ , rotated around the $y$ -axis.

$r=5\cos{\theta}$

Before we can plug into the formula, we need to find $x$ and $dr/d\theta$ .

Since $x=r\cos{\theta}$ , we get

$x=5\cos{\theta}\cos{\theta}$

$x=5\cos^2{\theta}$

To find $dr/d\theta$ , we’ll take the derivative of the given polar equation.

$r=5\cos{\theta}$

$\frac{dr}{d\theta}=-5\sin{\theta}$

We’ll plug everything into the formula for the surface area of revolution about the $y$ -axis.

$S_y=\int^{\beta}_{\alpha}2\pi{x}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\ d\theta$

$S_y=\int^{\pi}_02\pi\left(5\cos^2{\theta}\right)\sqrt{\left(5\cos{\theta}\right)^2+\left(-5\sin{\theta}\right)^2}\ d\theta$

$S_y=10\pi\int^{\pi}_0\cos^2{\theta}\sqrt{25\cos^2{\theta}+25\sin^2{\theta}}\ d\theta$

$S_y=10\pi\int^{\pi}_0\cos^2{\theta}\sqrt{25\left(\cos^2{\theta}+\sin^2{\theta}\right)}\ d\theta$

Since $\cos^2{\theta}+\sin^2{\theta}=1$ ,

$S_y=50\pi\int^{\pi}_0\cos^2{\theta}\sqrt{1}\ d\theta$

$S_y=50\pi\int^{\pi}_0\cos^2{\theta}\ d\theta$

Since $\cos^2{\theta}=\frac12\left[1+\cos{(2\theta)}\right]$ ,

$S_y=50\pi\int^{\pi}_0\frac12\left[1+\cos{(2\theta)}\right]\ d\theta$

$S_y=50\pi\int^{\pi}_0\frac12+\frac12\cos{(2\theta)}\ d\theta$

$S_y=25\pi\int^{\pi}_01+\cos{(2\theta)}\ d\theta$

$S_y=25\pi\left[\theta+\frac{1}{2}\sin{(2\theta)}\right]\bigg|^{\pi}_0$

$S_y=25\pi\theta+\frac{25\pi}{2}\sin{(2\theta)}\bigg|^{\pi}_0$

$S_y=25\pi(\pi)+\frac{25\pi}{2}\sin{(2\pi)}-\left[25\pi(0)+\frac{25\pi}{2}\sin{(2(0))}\right]$

$S_y=25\pi^2+\frac{25\pi}{2}(0)-25\pi(0)-\frac{25\pi}{2}(0)$

$S_y=25\pi^2$