# Volume of a Solid of Revolution

## Homework Statement

Find the volume generated by rotating the area bounded by $y^2 = 8x, x = 2$ and the x-axis about the y-axis.

## Homework Equations

Volume of cylinder, volume of disk.

## 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

## Homework Statement

Find the volume generated by rotating the area bounded by $y^2 = 8x, x = 2$ and the x-axis about the y-axis.

## Homework Equations

Volume of cylinder, volume of disk.

## 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

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

HallsofIvy
Homework Helper
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
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.

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.

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?

Thanks for your help!
Sheldon

Your integration is actually perfect.

Just notice that your answer is 1/2 the book's answer. That should lead you in the right direction.

Your integration is actually perfect.

Just notice that your answer is 1/2 the book's answer. That should lead you in the right direction.

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?

Thanks for your help,
Sheldon