- #1
kuahji
- 394
- 2
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=[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?
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?