# Integrating volume via shells

1. Feb 18, 2013

### PhizKid

1. The problem statement, all variables and given/known data
Volume of the region bounded by 1/x^4, y = 0, x = 2, and x = 6 about the axis y = -4

2. Relevant equations

3. The attempt at a solution

The height for any shell is x(y) - 2, or $\sqrt[4]{\frac{1}{y}} - 2$

The radius of any shell is y + 4

So the circumference is 2*pi*(y + 4), and the surface area is then [2*pi*(y + 4)] * $[\sqrt[4]{\frac{1}{y}} - 2]$

So I integrate this from 0 to 1/(2^4) because y(2) = 1/(2^4) which is the upper boundary on the y-axis and the lower boundary is 0.

Where did I go wrong?

2. Feb 18, 2013

### LCKurtz

The first place you went wrong was choosing to use shells in the first place. The problem is more naturally done with disks. Anyway, given that you are using shells, remember that the height of the shell in this example is $x_{right} - x_{left}$. The problem is that the right hand curve is not a single piece. Part of it is the line $x=6$. You have to break the integral into two parts to account for that.