# Volume of rotation - my answer seems off a bit

1. May 12, 2009

### lucidlobster

Find the area of the given function, rotated about the y axis

The Area below the line y=2 and above y=sin(x) from $$0-\pi$$

I did this rotated around the X AXIS no problem,

I found the area of the disk to be $$\pi(4-sin^{2}$$(x))

The volume around the X AXIS to be$$\frac{7\pi^{2}}{2}$$

3. The attempt at a solution

For the Y axis I have used this:$$\int^{\pi}_{0} 2\pi xf(x)dx$$
Plugging in:

$$\int^{\pi}_{0} 2\pi x(2-sin(x))dx$$

I solved this to be $$2\pi^{3}-2\pi^{2}$$

Does this look right?

2. May 12, 2009

### Staff: Mentor

Yes.

3. May 12, 2009

### lucidlobster

Really? That is awesome!

Any second opinions?

4. May 12, 2009

OK.
No.