The following solutions are for the introductory electrostatics problems linked here. I encourage you to attempt them by yourself first before looking through the solutions.

Problem 1

Using a method similar to our derivation of the washer method above, derive a similar formula to calculate the surface area of a solid of revolution.

Similar to the washer method for calculating volume, we will subdivide the solid of revolution as a stack of washers, each now with surface area \(2\pi rh\). In this case $r$ is $f(x)$ and $h$ is $dx$, and so the surface area $\mathcal{S}$ of a solid of revolution constructed by rotating the planar region between $y=f(x)$, the $x$-axis, $x=a$ and $x=b$ is given by

$$\mathcal{S}=2\pi\int_a^b f(x)dx$$

Of course, if $\mathcal{S}$ is a closed surface, then we also should add on the surface area contributions from the circular bases at $x=a$ and $x=b$:

$$\mathcal{S}=2\pi\int_a^b f(x)dx+\pi f(a)^2+\pi f(b)^2$$

Problem 2

Paul's Online Notes has an excellent set of practice problems (and associated solutions and answer explanations) that I encourage you to look over.