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January 30, 2015

January 30, 2015

Posted by **Elsi** on Saturday, December 15, 2012 at 1:17pm.

- Math - Calculus -
**Steve**, Saturday, December 15, 2012 at 4:27pmconsider 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

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