How to correctly calculate this integral using u substitution?

[ana111790]

Homework Statement

Calculate the following:
[PLAIN]http://img852.imageshack.us/img852/5580/integraltocalculate.jpg

Homework Equations

u substitution where: u= (1-r/R); du = -1/R dr

The Attempt at a Solution

= u_max*INT(-R*u^1/7 du)
= -u_max*R*(7/8)u^8/7 evaluated over 0,R
= -(7/8)*R*u_max [(1- R/R)^8/7 - (1- 0/R)^8/7]
* equals 0
**equals 1
= (7/8)*R*u_max

The answer at the end of the book says it should come out to be (49/60)*U_max (no R in the term)
Can anyone help me understand what I am doing wrong?

 

[SammyS]

Did you copy the problem correctly?

What you have looks fine.

Is this the complete problem, as stated?

 

[ana111790]

The integration is actually part of the following fluid mechanics problem:

Homework Statement

Water flows steadily through the round pipe in the figure. The entrance velocity is Uo. The exit velocity approximates turbulent flow, u = umax(1 − r/R)1/7. Determine the ratio
Uo/umax for this incompressible flow.
[PLAIN]http://img200.imageshack.us/img200/5879/fluidsproblem.jpg

Homework Equations

Conservation of mass

The Attempt at a Solution

Qin=Qout
V_o*A_in = A_out * INT(V(r) dr)
U₀*pi*r² = pi * r²* INT(u_max*(1- r/R)^1/7dr)

which is where the integration I wrote earlier comes in.
The final answer should be U₀/u_max = 49/60 but with my calculations I get U₀/u_max = 7R/8.

 

[SammyS]

u(r)=u_{max}\left(1-\frac{r}{R}\right)^{1/7} is the velocity of the fluid at distance, r, from the center of a pipe with circular cross-section. This makes sense, because the fluid at the center flows fastest, u(r) = u_MAX, and flow is slowest near the pipe surface: r → R, u(r) → 0.

The fluid is incompressible, so for an input flow velocity, u₀, the input flow rate, u₀·A_IN, is equal to the flow rate in the pipe and equal to the exit flow rate. To get the flow rate in the pipe and thus the exit flow rate, integrate u(r) over the cross-sectional area of the pipe.

u_0\cdot A_{in}=\int_0^Ru_{max}\left(1-\frac{r}{R}\right)^{1/7}{2\pi r}\,dr\ , where A_IN = πR².

See if this gets your desired answer.

It should.

To integrate, use the substitution, t=1-\frac{r}{R}\ then r=R(1-t)\,.

 

[ana111790]

I got the right answer now. Thank you so much for your help!