Thread: Volume calculation, function rotation around the x-axis.

1. 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

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

Have a nice day

2. 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...

3. 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.