# Thread: Work compute with integral

1. ## Work compute with integral

A tank is full of water. Find the work W required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3.
a = 5 b = 9 c = 12

So for this one, you have to set up an amount of rectangles.

The biggest rectangle is 5(or y) by 9, in relation to a smaller rectangle that is w by 9. I can also setup a relationship between the two to find that y/w=9/9 or y=w. So therefore we can produce an equation for the area of one rectangle.

dA=9y

Now the thickness of this rectangle is extremely small so the thickness is dy.

dV=9y dy

The density of water is 62.5 lb/ft^3, so this one rectangular-box of water is:

dM=562.5y dy

To find the force, we multiply by the force of gravity, I used 32.2 ft/sec^2.

dF=18112.5y dy

So our work is dF times the distance. This is where I become lost. Would the distance be vertical or horizontal? would it be (5-y) or (12-y)?

2. Well, after working the problem I came up with 50625 ft-lb. Wrong. Ugh.

I tried using a linear relationship. y=12/5y+12

dV=9(12/5y+12) dy
dV=108/5*y + 108

I multiplied by the density of water(62.5 lb/ft^3). d=m/v then d*v=m.

dV=1350y+6750 dy

Now I integrate from zero to five.

[675x^2+6750x](zero to five)=50625 ft-lb...

3. Anyone?

Thank you.

4. ^^ I just entered your answer into my homework checker and it told me I was wrong.

5. I am sorry, I didn't notice the spout. Let me get back. I am at work now and busy.

6. I think what we forgot to do was multiply by x.

By similar triangles, $\frac{5}{12}=\frac{5-x}{y}, \;\ y=\frac{12}{5}(5-x)$

$(62.5)(12/5)(9)=1350$

$1350\int_{0}^{5}x(5-x)dx=28125 \;\ ft/lbs$

7. Yes! Finally, with one try left I get it. Thanks Galactus. It's quite the problem.

8. That was correct then?. I thought so. I just forgot the x last time.

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# a tank is full of water. find the work w required to pump the water out of the spout

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