# Thread: Intergration problem im stuck on

1. ## 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.

[IMG]file:///C:/Users/Sami/AppData/Local/Temp/moz-screenshot.jpg[/IMG]

2. Just put $\displaystyle z=1-\frac{y^2}{b^2}.$

3. sorry to appear stupid but not quite sure what that means? use substitution?

4. 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.

5. ## 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

6. The answer you give is not correct for the integral you give! Was the integral supposed to be $\displaystyle \int_0^b y\sqrt{1- y^2/b^2} dy$?

If so then the substitution you were given before was probably $\displaystyle u= 1- y^2/b^2$. Show exactly how you tried that substitution.

7. sorry about starting a new post,

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

any help would be great

8. Originally Posted by samipoyo
sorry to appear stupid but not quite sure what that means? use substitution?
Have you been taught the substitution method?

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

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

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

9. yes but havent dont any real intergration in a couple years now, will try again now thanks

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

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

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

It would be a big help if you showed your working.
From HallsofIvy's observation, the integral probbaly should become $\displaystyle -\frac{b}{2} \int_{z = 1}^{z = 1 - b} z^{1/2} \, dz$.

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