Question

# Assume that water flowing through a pipe with a circular cross section flows most rapidly at...

Assume that water flowing through a pipe with a circular cross section flows most rapidly at the center of the pipe and least rapidly near the pipe walls (this might be reasonable because of friction between the water and the pipe walls). Assume that the water speed at the walls is half as great as the speed at the center, and assume that the decrease is linear (i.e. the graph of water speed versus radius is a straight line). If the radius of the pipe is 18.0 cm and if the volume flux of water is 8.00 liters per second, find the speed of the water at the center of the pipe. (One liter is 1000 cubic centimeters or 0.001 cubic meters). Hint: Draw the graph of speed versus radius, and then do the integral for the volume flux of water through the tube symbolically. Finally, use the given values to solve for the unknown water speed at the pipe center.

So the ?vdA = Q where Q is the volumetric flow rate.

The relationship between r and v is the slope 1-r/2*(R) where R is 0.18 m.
So you'll get something like:
?[0,2*pi],[0,.18] of v(1-r/.2)rdrd(theta) = .008

v is technically a constant that you would multiply to get your velocity and is therefore the thing you want to solve for since v will be it's greatest in the center anyway. Oh and theta has no angular dependence so we can take that out.

2*pi*v* ?[0,.18] of r-r^2/.20 dr = .008

v= 3.654 e-1 m/s