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.(adsbygoogle = window.adsbygoogle || []).push({});

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=[tex]\sqrt{(16-y^2)}[/tex]

dV=40[tex]\sqrt{(16-y^2)}[/tex]

F(y)=57(40)[tex]\sqrt{(16-y^2)}[/tex]

10-y should be the distance the work must do

W=2280 [tex]\int[/tex](10-y)[tex]\sqrt{(16-y^2)}[/tex]

Then I distributed the (10-y)

W=22800[tex]\int[/tex][tex]\sqrt{(16-y^2)}[/tex]- 2280[tex]\int[/tex]y[tex]\sqrt{(16-y^2)}[/tex]

For part two, I set u=16-y^2 & got

W=22800[tex]\int[/tex][tex]\sqrt{(16-y^2)}[/tex]+ 1140[tex]\int[/tex]y[tex]\sqrt{(16-y^2)}[/tex]

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[tex]\pi[/tex] as follows

22800(4[tex]\pi[/tex])+ 1140[tex]\int[/tex]y[tex]\sqrt{(16-y^2)}[/tex] (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?

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# Homework Help: Pumping a cylindrical tank

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