# Pumping a cylindrical tank

• December 5th 2007, 03:16 PM
kuahji
Pumping a cylindrical tank
A storage tank is a right circular cylinder 20ft long and 8 ft in diameter with its axis horizontal. If the tank is half full of olive oil weighing 57 lb/ft^3, find the work done in emptying it through a pipe that runs from the bottom of the tank to an outlet that is 6 ft above the top of the tank.
My work
I set the bottom of the tank at the point (0,0) & then drew a circle with a radius of 4.
dV=2x(20)dy
then solved the the circle's equation for x, x= $\sqrt{(16-y^2)}$
dV=40 $\sqrt{(16-y^2)}$
F(y)=57(40) $\sqrt{(16-y^2)}$
10-y should be the distance the work must do
W=2280 $\int$(10-y) $\sqrt{(16-y^2)}$
Then I distributed the (10-y)
W=22800 $\int$ $\sqrt{(16-y^2)}$- 2280 $\int$y $\sqrt{(16-y^2)}$
For part two, I set u=16-y^2 & got
W=22800 $\int$ $\sqrt{(16-y^2)}$+ 1140 $\int$y $\sqrt{(16-y^2)}$
This is the part where I get lost, I did it a bit differently from the solutions manual, but at this point the manual shows inserting 4 $\pi$ as follows
22800(4 $\pi$)+ 1140 $\int$y $\sqrt{(16-y^2)}$ (evaluated from 0 to -4), I tried from (0 to 4) in my solution.
I don't understand why the solutions manual is doing this particular step, it says its the area of a semicircle, but can anyone explain why I use it in this problem & where the integral disappears to?
• December 6th 2007, 05:38 AM
kalagota
Quote:

Originally Posted by kuahji
A storage tank is a right circular cylinder 20ft long and 8 ft in diameter with its axis horizontal. If the tank is half full of olive oil weighing 57 lb/ft^3, find the work done in emptying it through a pipe that runs from the bottom of the tank to an outlet that is 6 ft above the top of the tank.
...

This is the part where I get lost, I did it a bit differently from the solutions manual, but at this point the manual shows inserting 4 $\pi$ as follows
22800(4 $\pi$)+ 1140 $\int$y $\sqrt{(16-y^2)}$ (evaluated from 0 to -4), I tried from (0 to 4) in my solution.
I don't understand why the solutions manual is doing this particular step, it says its the area of a semicircle, but can anyone explain why I use it in this problem & where the integral disappears to?

in fact, it was the area of a quarter circle:
note that the "function" was $g(y) = \sqrt{16 - y^2}$, which is actually a semicircle on the right side of the y-axis.. but since it was only integrated from -4 to 0, then the integral was the "area" of the region of the circle on the fourth quadrant only.. and since the whole circle is of radius 4, then $A_{circle} = \pi r^2 = 16\pi$ and you only need the quarter of it and therefore, $\frac{A_{circle}}{4} = 4\pi$