# Homework Help: Finding volume of axis of revolution

1. Aug 25, 2010

### Sidthewall

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

$$2\pi \int_0^6 y\sqrt{25- (y- 1)^2}dy + 5\pi$$
That's the integral i need solved

2. The attempt at a solution[/b]
so first i subbed u=y-1
took the 2 pi out of the integral
that got me 2 integrals u*sqrt(25-(u)^2) du + sqrt(25-(u)^2) du
the first integral = (2*pi*24^(3/2))/3

the second i used a trig sub and my final answer is
(25pi^2)/2 +
-25*asin(-.2)+sqrt(24)

The final answer is suppose to be 25pi^2 + 500pi/3

but i got25pi^2 + (-75*asin(-.2)+3*sqrt(24)+2*(24)^(3/2) -15)/3
which is approx
25pi^2 + 499.89pi/3

how do i get the exact answer

2. Aug 25, 2010

### Staff: Mentor

I infer from the problem title that you are rotating some curve around some axis. Are you sure that your integral is the right one for your problem? I don't want to spend any time with the integral you have if I'm not sure it represents the problem you're working.

3. Aug 25, 2010

### Sidthewall

Yes I am 100% sure I am using the right integrel. The integral itself is odd and a longer and harder way to find the volume of this object. I would normally solve this in respect to x but the instrucotr wants it in respect to y.
And yeah I am 100% sure it is the right integrel

4. Aug 26, 2010

### Staff: Mentor

That looks fine.
25pi/4 - 25/2 * sin(2arcsin(-1/5)) - 25/2 * arcsin(-1/5).

I don't have time to check that before going to work, so I might have made a mistake.