Find the volume of a function that revolves by pi radians.

Question

I know that the volume of a function which revolves around an axis by a full turn is the following: $$V=\pi \int_a^b f^2(x)dx$$ But I'm confused as to what to do when the function is asked to be rotated by an amount of π (such as π/2 or 2π or just π). Is a full turn equal to π when it comes to that formula? If not why isn't 2π used in the formula instead? I'd really like to clean this up in my head.

Answer 1

The formula you wrote can be seen as a generalization of the formula for the volume of a cylinder $-$ that's why there's a $\pi$ not a $2\pi$: $$V_{\text{cylinder}} = \pi r^2 h,$$ where $r\geq 0$ is the radius and $h\geq 0$ is the height of the cylinder $-$ if we think of the cylinder standing upright in three-dimensional space, this is the $z$-coordinate.

Indeed, assume that the radius is dependent on the $z$-coordinate (as is the case with a rotationally symmetric object), i.e. if $z\in[0,h]$ is the $z$-coordinate, then the volume is given by $$\int\limits_0^h \pi r(z)^2dz.$$

So, coming back to the cylinder: If one only takes $\frac{1}{2}$ of its basis disc into account, the formula becomes $\frac{1}{2} \pi r^2h$, and analogously with any ratio.

This generalizes to the volume formula.