# Washer method problem

1. Nov 19, 2008

### elitespart

1. Find the volume generated by revolving the area bounded by x=y$$^{2}$$ and x=4 about the line y=2.

Is it $$\pi\int(4-(2-\sqrt{x})^{2}dx$$ from 0 to 4?

2. Nov 19, 2008

### Dick

No. The inner radius is (2-sqrt(x)). What's the outer radius? It doesn't lie on the x-axis.

3. Nov 19, 2008

### elitespart

oh whoops forgot to add a bracket at the end. Thanks for your help.

4. Nov 19, 2008

### Dick

Hold it. I gave it a second thought and edited my answer. The area enclosed by x=4 and y^2=x has pieces both about and below the x-axis.

Last edited: Nov 19, 2008
5. Nov 19, 2008

### elitespart

So it's 4 instead of 2?

6. Nov 19, 2008

### Dick

No, y^2=x gives y=sqrt(x) or y=-sqrt(x). Those are the boundaries of the region. Your original answer would be ok if they had specified y=0 as a boundary. But they didn't.

7. Nov 19, 2008

### elitespart

oh okay. So would y=2 factor into the outer radius too? Meaning r = [2 - (-sqrt(x))]?

8. Nov 19, 2008

### Dick

Exactly. The outer radius is 2+sqrt(x).

9. Nov 19, 2008

### elitespart

awesome. Thanks again.