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

1. Aug 14, 2011

### flyingpig

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution

Which one is right?

$$\int_{0}^{1} \pi (x^2 - \sqrt{x})^2 dx$$

$$\pi \int_{0}^{1} x^4 - x dx$$

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?

2. Aug 14, 2011

### Quinzio

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

3. Aug 14, 2011

### 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.