Show that $2\pi \int_{\alpha}^{\beta}yg(y)dy=b\pi\beta^2 - a\pi\alpha^2-\pi\int_a^b[f(x)]^2dx$

Question

Let $f(x)$ and $g(y)$ be inverse functions with $f(a)=\alpha$, $f(b)=\beta$, $0\le a\lt b$, $0\le \alpha \lt \beta$. Show that $$2\pi \int_{\alpha}^{\beta}yg(y)dy=b\pi\beta^2 - a\pi\alpha^2-\pi\int_a^b[f(x)]^2dx$$

Volume of a solid of revolution: Disk method $$\pi\int_a^b[f(x)]^2dx$$

Volume of a solid of revolution: Shell method $$2\pi\int_a^bxf(x)dx$$

As far as I can tell, this assumption is giving and answer of the shell method in terms of the disk method of the inverse function, but still don’t know how to prove this assumption.

I’d appreciate any help. Thank you.

Answer 1

Integrate-by-parts the LHS,

$$2\pi \int_{\alpha}^{\beta}yg(y)dy = \pi\int_{\alpha}^{\beta}g(y)d(y^2) = \pi (I_1 - I_2) \tag 1$$

where

$$I_1 = g(y)y^2 \bigg|_{\alpha}^{\beta}=g(\beta)\beta^2-g(\alpha)\alpha^2 = g(f(b))\beta^2-g(f(a))\alpha^2 =b\beta^2-a\alpha^2\tag 2$$

and, with the variable change $y=f(x)$ for $I_2$,

$$I_2 = \int_{\alpha}^{\beta}y^2g'(y)dy = \int_{a}^{b} [f(x)]^2 g'(f(x))f'(x)dx= \int_{a}^{b} [f(x)]^2 dx\tag 3$$

Note that $g'(f(x))f'(x)=1$, derived from $[g(f(x)]' = x'=1$ is used in the last step. Plug (2) and (3) into (1) to arrive at

$$2\pi \int_{\alpha}^{\beta}yg(y)dy=b\pi\beta^2 - a\pi\alpha^2-\pi\int_a^b[f(x)]^2dx$$