# Thread: Lost on Question 18...What am I doing wrong???

1. ## Lost on Question 18...What am I doing wrong???

Example: The conical tank from y = 2x is filled to within 2 ft of the top with the olive oil weighing 57 lb/ft^3. How much work does it take to pump the oil to the rim of the tank.

Delta V = Pi (radius)^2 (thickness) = Pi(1/2y)^2 delta y = pi/4 y^2delta y ft^3
F(y) = 57 delta V = 57 Pi/4 y^2 delta y lb.
Delta W = 57pi/4 (10-y) y^2
W = Integral from 0 to 8 [ 57pi/4 (10-y)y^2dy ~ 30,561 ft-lb

Now the problem: Supposed that, instead of being full , the tank in Example 5 is only half full. How much work does it take to pump the remaining oil to a level 4 feet above the top of the tank? Answer: 60,042 ft-lb.

ok my process

For problem 18
Since it's 1/2 the volume capacity...I took half of the volume of Example 3 and the integral changes and kept everything else the same....
What's wrong?
V = 1/2[(pi)(1/2y)^2(delta y)] = 1/2 (Pi/4y^2delta y) ft^3
F(y) = (57 lb/ft^3) 1/2[pi/4 y^2 delta y] = 57/2 [pi/4 y^2 delta y]
W = 57/2 [integral from 0 to 6 pi/4 y^2 (10-y) dy = 57pi/8 [integral from 0 to 6 y^2(10-y))
= 57pi/8 [integral from 0 to 6 10y^2-y^3] = 57pi/8[10y^3/3-y^4/4] from 0 to 6
= 57pi/8 [10(6)^3/3 - (6)^4/4] = 8,864 ft lb

2. Originally Posted by Nimmy
Example: The conical tank from y = 2x is filled to within 2 ft of the top with the olive oil weighing 57 lb/ft^3. How much work does it take to pump the oil to the rim of the tank.

Delta V = Pi (radius)^2 (thickness) = Pi(1/2y)^2 delta y = pi/4 y^2delta y ft^3
F(y) = 57 delta V = 57 Pi/4 y^2 delta y lb.
Delta W = 57pi/4 (10-y) y^2
W = Integral from 0 to 8 [ 57pi/4 (10-y)y^2dy ~ 30,561 ft-lb

Now the problem: Supposed that, instead of being full , the tank in Example 5 is only half full. How much work does it take to pump the remaining oil to a level 4 feet above the top of the tank? Answer: 60,042 ft-lb.
Where did you get this? Doesn't it seem strange that pumping half the oil would take nearly twice the work?

ok my process

For problem 18
Since it's 1/2 the volume capacity...I took half of the volume of Example 3 and the integral changes and kept everything else the same....
What's wrong?
V = 1/2[(pi)(1/2y)^2(delta y)] = 1/2 (Pi/4y^2delta y) ft^3
F(y) = (57 lb/ft^3) 1/2[pi/4 y^2 delta y] = 57/2 [pi/4 y^2 delta y]
W = 57/2 [integral from 0 to 6 pi/4 y^2 (10-y) dy = 57pi/8 [integral from 0 to 6 y^2(10-y))
= 57pi/8 [integral from 0 to 6 10y^2-y^3] = 57pi/8[10y^3/3-y^4/4] from 0 to 6
= 57pi/8 [10(6)^3/3 - (6)^4/4] = 8,864 ft lb

3. Well my prof...told me that....

For #18: You're right that the volume is half of what it was before, but
the 1/2 doesn't play into the V equation (which is the one you use for the
volume of a "slab."). Instead, use the fact that we have half the initial
volume to find the new height of the oil in the cone. This will make your
work formula a little different.

I got the answer from the professor.

Originally Posted by HallsofIvy
Where did you get this? Doesn't it seem strange that pumping half the oil would take nearly twice the work?
ok my process

For problem 18
Since it's 1/2 the volume capacity...I took half of the volume of Example 3 and the integral changes and kept everything else the same....
What's wrong?
V = 1/2[(pi)(1/2y)^2(delta y)] = 1/2 (Pi/4y^2delta y) ft^3
F(y) = (57 lb/ft^3) 1/2[pi/4 y^2 delta y] = 57/2 [pi/4 y^2 delta y]
W = 57/2 [integral from 0 to 6 pi/4 y^2 (10-y) dy = 57pi/8 [integral from 0 to 6 y^2(10-y))
= 57pi/8 [integral from 0 to 6 10y^2-y^3] = 57pi/8[10y^3/3-y^4/4] from 0 to 6
= 57pi/8 [10(6)^3/3 - (6)^4/4] = 8,864 ft lb[/QUOTE][/QUOTE]