Calculating Volume of a Cylinder Using Integration

[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 haven't found a substitution that would get rid of my R variable

 

[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.

 

[jennypear]

thanks so much!