# Volume of solid x=1+y^2 using shell method

1. May 25, 2013

### Painguy

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

Volume of solid rotated about x-axis x=1+y^2, y=1, y=0, x=0 using shell method

2. Relevant equations

3. The attempt at a solution

so i set up the integral with

∫2pi(y)(1+y^2)dy from 1 to 2

which is apparently wrong, but i don't know why.

2. May 25, 2013

### haruspex

You wrote y=0 as a bound, but I guess you meant y=2.
The formula looks right. What did you get? Do you know what the 'right' answer is?

3. May 25, 2013

### Painguy

Oh right. My bad I did mean y=2. Well here is my professors solution. If evaluate his answer I get 14. If I evaluate mine I get around 32.

Here is his solution. He seems to be using the washer method.

4. May 25, 2013

### Skins

Your work via the shell method is correct. The result computes to the same value using the disc method that your professor used and the shell method you used.. But it's easier to do it using shells because you only have to deal with 1 integral. Go back and check your computation. You should be getting the same result via both methods. Neither of the final results you got seem to be correct despite the fact that the integrals are correct.

Last edited: May 25, 2013
5. May 25, 2013

### Painguy

I tried again. This time my professor's answer came out to be around 80.11. My answer however remained at 32, and wolfram alpha and my calculator give the same answer. here is my work.

6. May 25, 2013

### haruspex

80 is obviously too much. The whole solid fits inside a rectangular block 4x4x5. I agree with approx 32 (21 pi/2).

7. May 26, 2013

### Painguy

Did I evaluate my professors integral incorrectly?

8. May 26, 2013

### haruspex

Your Prof wrote √(1-x), but it should have been √(x-1).

9. May 26, 2013

### Skins

I get the result

$$\frac{21 \pi}{2}$$

via both shell (cylinder) and disc methods)

As I said, the integral you wrote down is correct but somewhere you must have messed up your computation.