# Volume of a Solid of Revolution

1. Apr 8, 2007

### SheldonG

1. The problem statement, all variables and given/known data
Find the volume generated by rotating the area bounded by $y^2 = 8x, x = 2$ and the x-axis about the y-axis.

2. Relevant equations
Volume of cylinder, volume of disk.

3. The attempt at a solution
I think this can be solved by subtracting the 'empty' volume from the volume of a cylinder of radius 2:

$$\pi r^2h - \int_0^4 \frac{\pi y^2}{8}\,dy$$

I get an answer of $\pi 40/3$ but the solution provided is $128/5\pi$.

I'm sure my arithmetic is not that far off :-)

What did I do wrong?

Thanks,
Sheldon

2. Apr 8, 2007

### SheldonG

I see an error, but I still don't get the provided answer. I believe the following is correct:

$$\pi r^2h - \int_0^4 \frac{\pi y^4}{64}\,dy$$

However I get the answer: $64\pi/5$

Thanks!
Sheldon

3. Apr 8, 2007

### HallsofIvy

Staff Emeritus
That integral is the correct one. I don't see how you could get $64\pi/5$ out of it! Show exactly how you integrated.

4. Apr 8, 2007

### SheldonG

When x = 2, y = 4, so that's the height of the cylinder:

$$\pi(2^2)4 - \int_0^4 \frac{\pi y^4}{64}\,dy = 16\pi -\pi\left[\frac{y^5}{(5)(64)}\right]_0^4$$

$$16\pi - \frac{16\pi}{5} = \frac{80\pi}{5} - \frac{16\pi}{5} = \frac{64\pi}{5}$$

Am I making some dumb mistake here?

Sheldon

5. Apr 8, 2007

6. Apr 8, 2007

### SheldonG

Hi Wolf,

I noticed that, but I don't understand it. I am starting to wonder if the book answer is wrong, but this is a new topic, and I am not feeling very confident.

If I multiply by two (or integrate from -4 to 4), then I get the book answer. But the problem states that the area is bounded by the X-axis. Multiplying by 2 would give the volume for the object both above and below the X-axis. Isn't that right?