# Math prob

1. Jul 20, 2005

### jennypear

v=(1/pi*Ro) integral 20(1-R/Ro)^(1/7) 2*pi*R*dR
i know that i need to change my variable

started out y=Ro-R
dy=-dR

but havent found a substitution that would get rid of my R variable

2. Jul 20, 2005

### HallsofIvy

?? R0- R certainly should "get rid of" the R variable. I'm not completely certain whether that first R0 is in the denominator with pi or not. I'll assume it's not.
What you have is $$\frac{20}{\pi}R_0(R_0)^{-\frac{1}{7}}(2\pi)\int(R_0-R)^{\frac{1}{7}}RdR$$$$=40R_0^{\frac{6}{7}}\int(R_0-R)^{\frac{1}{7}}RdR$$.
Let y= R0- R so that dy= -dR and R= R0- y. Then the integral becomes $$-40R_0^{\frac{6}{7}}\int y^{\frac{1}{7}}(R_0-y)dy= -40R_0^{\frac{6}{7}}\int(R_0y^{\frac{1}{7}}- y^{\frac{8}{7}})dy$$ which is easy.

Last edited by a moderator: Jul 20, 2005
3. Jul 20, 2005

### jennypear

thanks so much!

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