# Volume calculation, function rotation around the x-axis.

• May 11th 2011, 02:09 AM
helpwouldbenice
Volume calculation, function rotation around the x-axis.
Hello everyone.

I'm in need of some help with Volume calculations (with a function that is rotating around the x-axis).

I got stuck on the following: y = 3/sqrt{2-x} interval: -1 =< x =< 1

{1 {1
\pi \int y^2 = \pi \int 9/(2-x) = /pi [Need help with this bit]
{-1 {-1

I think my big problem is to find correct Antiderivative.

Did edit a little to make it more understandable.
Sorry but i dont understand how latex work (so it looks ugly i know) and the help page on this site doesn't work (Shake)

I would be very thankful if anyone with more insight than me could give me some help. Primarily with Antiderivative.

(Hi) Have a nice day
• May 11th 2011, 02:32 AM
Prove It
You're right so far, except the bounds of your integral are -1 and 1, and you need to write y in terms of x...
• May 11th 2011, 01:25 PM
Prove It
Now that you've got the integral to $\displaystyle \int_{-1}^1{\frac{9\pi}{2 - x}\,dx} = 9\pi\int_{-1}^1{\frac{1}{2-x}\,dx}$, solve using a u substitution.