Find, with calculus shapes of function $y=f(x)$, $y = g(x)$

Question

There's a question of integrals. The region between some unknown function, $y=f(x)$,$y = g(x)$, and $f(x) \geq g(x) \geq 0$ in intervals $[2,5]$ is rotated about the lines, $x=9$ to make a 3D shape. So how do I find the volume of the shape? I personlly use disk and sphere method, and found $ \int_{2}^{5} \pi (f(x)+9)^2-(g(x)+9)^{2}dx$ is correct answer. So I'm here to ask questions to check if this is correct, or this is wrong. But if it's wrong answer, then please tell me if which one is right.

Answer 1

Your formula is not correct and since you didn't show how you found it, we cannot check where you made mistakes.

Note, that for $x\in [2,5]$, the distance to the axis of rotation $x=9$ is $r(x) = 9-x$. Rotating about the given axis gives rise to a cylinder with infinitesimal thickness $dx$ and a height of $h(x) = f(x)-g(x)$.

The Volume of such an infinitesimal cylinder is

$$dV(x) = 2\pi r(x)h(x)dx\Rightarrow V =2\pi\int_2^5(9-x)(f(x)-g(x))\;dx$$