# Determining the Surface Area of a Solid of Revolution

## Key Questions

• Write :

$r = 3 \sin \left(\theta\right)$

$r = 3 \frac{y}{r}$ because $y = r \sin \left(t\right)$

${r}^{2} = 3 y$

${x}^{2} + {y}^{2} = 3 y$ because $r = \sqrt{{x}^{2} + {y}^{2}}$.

${x}^{2} + {\left(y - \frac{3}{2}\right)}^{2} = \frac{9}{4}$

You recognize a circle of radius $\frac{3}{2}$. The area is $\pi {\left(\frac{3}{2}\right)}^{2} = 9 \frac{\pi}{4}$.

• If a surface is obtained by rotating about the x-axis the polar curve $r = r \left(\theta\right)$ from $\theta = {\theta}_{1}$ to ${\theta}_{2}$, then its surface area A can by found by

$A = 2 \pi {\int}_{{\theta}_{1}}^{{\theta}_{2}} r \left(\theta\right) \sin \theta \sqrt{{r}^{2} + {\left[r ' \left(\theta\right)\right]}^{2}} d \theta$.

I hope that this was helpful.

## Questions

There are no questions in this topic. See questions for all topics.