- #1

- 16

- 0

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

- Thread starter jennypear
- Start date

- #1

- 16

- 0

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

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

?? 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 [tex]\frac{20}{\pi}R_0(R_0)^{-\frac{1}{7}}(2\pi)\int(R_0-R)^{\frac{1}{7}}RdR[/tex][tex]=40R_0^{\frac{6}{7}}\int(R_0-R)^{\frac{1}{7}}RdR[/tex].

Let y= R_{0}- R so that dy= -dR and R= R_{0}- y. Then the integral becomes [tex]-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[/tex] which is easy.

What you have is [tex]\frac{20}{\pi}R_0(R_0)^{-\frac{1}{7}}(2\pi)\int(R_0-R)^{\frac{1}{7}}RdR[/tex][tex]=40R_0^{\frac{6}{7}}\int(R_0-R)^{\frac{1}{7}}RdR[/tex].

Let y= R

Last edited by a moderator:

- #3

- 16

- 0

thanks so much!

- Last Post

- Replies
- 5

- Views
- 1K

- Replies
- 3

- Views
- 2K

- Replies
- 2

- Views
- 8K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 3K

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 746

- Last Post

- Replies
- 13

- Views
- 2K

- Last Post

- Replies
- 7

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 1K