Solids of Revolution

Concept:

Imagine you had a function $f(x)=\sqrt{x}$, with domain $[0,3]$. Visualise that. Now, imagine the lower end is stuck on a swivel and you pluck the top end and rotate it around the x-axis, leaving a wall as it passes. The solid you just created is a solid of revolution.

Finding A(x)

Now the fun part. How do you calculate the volume of it? First, what’s the $A(x)$ (the area) of a cross section of the shape? Well, it’s a circle, the radius is $\sqrt{x}$. $\therefore A(x)=\pi (\sqrt{x}{)}^2=\pi x$.

Finding Volume:

Using Volume by Slicing, we can pretty easily find it now. Just integrate, plugging in $A(x)$ instead of just $x$.

What about harder shapes?

Well most of the time you won’t get something as simple as a single function around the x-axis (which gives you a nice and convenient cross sectional area). Instead, you’ll more often be asked to calculate the volume of a solid that was created from revolution of the area enclosed by 2 functions (take $f(x)=x$ and $g(x)=x^2$. This looks like the below. The area enclosed is called a washer:

Thus the formula for the Cross Sectional Area (applied to the question from screenshot): $A(x)=\pi r_{outer}^2-\pi r_{inner}^2=\pi (x{)}^2-\pi ((x{)}^2{)}^2=\pi (x^2-x^4)$