Integrating Volume via Shells: What Went Wrong with the Bounds?

[PhizKid]

Homework Statement

Volume of the region bounded by 1/x^4, y = 0, x = 2, and x = 6 about the axis y = -4

Homework Equations

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?

 

[LCKurtz]

Homework Statement

Volume of the region bounded by 1/x^4, y = 0, x = 2, and x = 6 about the axis y = -4

Homework Equations

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?

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.