I can't believe I am not getting this, shell method

[flyingpig]

Homework Statement

Let's say I have the area bounded by [tex]y_1 = \sqrt{x}[/tex] and [tex]y_2 = x^2[/tex] in (0,1). Rotate that about the x-axis, find that volume of solid.

The Attempt at a Solution

Which one is right?

[tex]\int_{0}^{1} \pi (x^2 - \sqrt{x})^2 dx[/tex]

[tex]\pi \int_{0}^{1} x^4 - x dx[/tex]

I think the second integral is right because it uses washeres?

If the second integral is wrong, what is the meaning of the first integral?

 

[Quinzio]

The second is the correct one.
The first maybe be... something like a "cone" shaped solid whose radious is [itex]y_1(x)-y_2(x)[/itex]

 

[doppelganger]

I think you have the signs of the terms in the second integral reversed. The square root of x is larger than x squared on the interval from 0 to 1. If you evaluate your integral, you get a negative volume.