# Volume of a solid

• May 15th 2011, 08:55 PM
jackie
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 $\displaystyle y=f(x)$ where $\displaystyle 0\leqslant x \leqslant c$ about the x-axis, and the resulting volume is $\displaystyle c^2 + c$. We want to find the function $\displaystyle f(x)$.
We have the volume using disk method $\displaystyle V=\int_0^c \pi r^2 dx=c^2+c$. But $\displaystyle r=y=f(x)$, and I don't know how to change the variable of the expression inside the integral so that I can integrate it.
• May 15th 2011, 10:03 PM
earboth
Quote:

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 $\displaystyle y=f(x)$ where $\displaystyle 0\leqslant x \leqslant c$ about the x-axis, and the resulting volume is $\displaystyle c^2 + c$. We want to find the function $\displaystyle f(x)$.
We have the volume using disk method $\displaystyle V=\int_0^c \pi r^2 dx=c^2+c$. But $\displaystyle r=y=f(x)$, 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:

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

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

2. Now calculate backwards:

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

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

Change variables and differentiate both sides:

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