# Math Help - Volume of a solid

1. ## Volume of a solid

My friend and I ran into another problem that we were not able to solve. If anyone can gives us a hint, we'd really appreciate it.

A solid is generated by rotating the curve $y=f(x)$ where $0\leqslant x \leqslant c$ about the x-axis, and the resulting volume is $c^2 + c$. We want to find the function $f(x)$.
We have the volume using disk method $V=\int_0^c \pi r^2 dx=c^2+c$. But $r=y=f(x)$, and I don't know how to change the variable of the expression inside the integral so that I can integrate it.

2. Originally Posted by jackie
My friend and I ran into another problem that we were not able to solve. If anyone can gives us a hint, we'd really appreciate it.

A solid is generated by rotating the curve $y=f(x)$ where $0\leqslant x \leqslant c$ about the x-axis, and the resulting volume is $c^2 + c$. We want to find the function $f(x)$.
We have the volume using disk method $V=\int_0^c \pi r^2 dx=c^2+c$. But $r=y=f(x)$, and I don't know how to change the variable of the expression inside the integral so that I can integrate it.
1. Rotation about the x-axis is calculated by:

$V = \int\left( \pi \cdot (f(x))^2\right) dx$

That means the r in your equation correspond to the function f.

2. Now calculate backwards:

$V = \int_0^c \left( \pi \cdot (f(x))^2\right) dx = c^2+c$

$\int_0^c \left( (f(x))^2\right) dx = \dfrac1{\pi} (c^2+c)$

Change variables and differentiate both sides:

$(f(x))^2 = \dfrac2{\pi} x + \dfrac1{\pi}~\implies~f(x) = \sqrt{\dfrac2{\pi} x + \dfrac1{\pi}}$