consider the amount of water that flows through a cross-section of the pipe in 1 second. It is a conical solid, high in the center and zero at the edges.
At radius r, the volume of the cylindrical shell of water is 2pi*r*h dr, where h = 10(1-r^2)
So, the amount across the whole cross-section is
integral[0,1] 2pi*r*10(1-r^2) dr
= 10pi r^2 - 5pi r^4 [0,1]
= 5pi in^3
since that's the volume in 1 second, the rate is 5pi in^3/s
Water is flowing in a cylindrical pipe of radius 1 inch. Because water is viscous and sticks to the pipe, the rate of flow varies with distance from the center. The speed of the water at distance r inches from the center is 10(1-(r^2)) per second. What is the rate (in cubic inches per second) at which water is flowing through the pipe?
1 answer
1 answer