# Intergration problem im stuck on

• Oct 22nd 2008, 11:00 AM
samipoyo
Intergration problem im stuck on
I stuck in a engineering question at some intergration any help in how to do it rather than just the answer would be very helpfull as i have the asnwer but have no idea where it comes from.

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• Oct 22nd 2008, 11:06 AM
Krizalid
Just put $z=1-\frac{y^2}{b^2}.$
• Oct 23rd 2008, 02:26 AM
samipoyo
sorry to appear stupid but not quite sure what that means? use substitution?
• Oct 23rd 2008, 03:53 AM
mr fantastic
Quote:

Originally Posted by samipoyo
sorry to appear stupid but not quite sure what that means? use substitution?

Yes. Use the substitution that has been suggested.
• Oct 25th 2008, 05:31 AM
samipoyo
Same intergration, still struggling
I have recently posted a intergration problem and with a substitutional hint i have retried and refailed, please can someone give me a few steps to help me as i simply dont want to use the answer as i will have achieved nothing
• Oct 25th 2008, 05:36 AM
HallsofIvy
The answer you give is not correct for the integral you give! Was the integral supposed to be $\int_0^b y\sqrt{1- y^2/b^2} dy$?

If so then the substitution you were given before was probably $u= 1- y^2/b^2$. Show exactly how you tried that substitution.
• Oct 25th 2008, 05:38 AM
samipoyo
sorry about starting a new post,

I sub in and end up with another intergration i cant do

any help would be great
• Oct 25th 2008, 05:39 AM
mr fantastic
Quote:

Originally Posted by samipoyo
sorry to appear stupid but not quite sure what that means? use substitution?

Have you been taught the substitution method?

$z = 1 - \frac{y^2}{b} \Rightarrow \frac{dz}{dy} = - \frac{2y}{b} \Rightarrow dy = -\frac{b}{2y} \, dz$.

The integral becomes $-\frac{b}{2} \int_{z = 1}^{z = 1 - b} z \, dz$.

It would be a big help if you showed your working.
• Oct 25th 2008, 05:42 AM
samipoyo
yes but havent dont any real intergration in a couple years now, will try again now thanks
• Oct 25th 2008, 05:59 AM
mr fantastic
Quote:

Originally Posted by mr fantastic
Have you been taught the substitution method?

$z = 1 - \frac{y^2}{b} \Rightarrow \frac{dz}{dy} = - \frac{2y}{b} \Rightarrow dy = -\frac{b}{2y} \, dz$.

The integral becomes $-\frac{b}{2} \int_{z = 1}^{z = 1 - b} z \, dz$.

It would be a big help if you showed your working.

From HallsofIvy's observation, the integral probbaly should become $-\frac{b}{2} \int_{z = 1}^{z = 1 - b} z^{1/2} \, dz$.

It would help if you confirmed what the actual integral is ....