Calculating work done in draining water

[doctordiddy]

Homework Statement

http://imgur.com/BWo3tLh

Using the fact that the weight of water is 62.5lb/ft3, find the work (in ft-lbs) required to pump the water out of the outlet when r =3ft. Make sure your answer is correct to within ten ft-lbs.

Homework Equations

The Attempt at a Solution

I figured that Work is just force*distance, and force in this case is 62.5*V, while distance is just 3-height of the water, which i set as y

I then get this equation for work

62.5V(3-y)

to find volume i wrote this as

∫62.5(pi)r^2(3-y) from y=-3 to y=0 bottom of hemisphere to top)which can be written as

62.5pi∫r^2(3-y)

I then tried to figure out a relationship between r and y, and i figured that because the hemisphere fit on the graph y=√(9-x^2) and then i plugged in r for x and rearranged it to give

r=√(9-y^2)

Then i put this back into the equation to finally get

62.5pi∫(9-y^2)(3-y) = 62.5pi∫27-9y-3y^2+y^3

After integrating i get 62.5pi(27y-9y^2/2-y^3+y^4/4)

Then plugging in the points from -3 to 0 i finally end up with

-62.5pi(-81-81/2+27+81/4) =~14579 ft lbs

this is incorrect though.

I feel like i may have used a wrong substitution for r, but i do not know how i can relate r and y. can anyone help?

Thanks

 

[Dick]

Homework Statement

http://imgur.com/BWo3tLh

Using the fact that the weight of water is 62.5lb/ft3, find the work (in ft-lbs) required to pump the water out of the outlet when r =3ft. Make sure your answer is correct to within ten ft-lbs.

Homework Equations

The Attempt at a Solution

I figured that Work is just force*distance, and force in this case is 62.5*V, while distance is just 3-height of the water, which i set as y

I then get this equation for work

62.5V(3-y)

to find volume i wrote this as

∫62.5(pi)r^2(3-y) from y=-3 to y=0 bottom of hemisphere to top)which can be written as

62.5pi∫r^2(3-y)

I then tried to figure out a relationship between r and y, and i figured that because the hemisphere fit on the graph y=√(9-x^2) and then i plugged in r for x and rearranged it to give

r=√(9-y^2)

Then i put this back into the equation to finally get

62.5pi∫(9-y^2)(3-y) = 62.5pi∫27-9y-3y^2+y^3

After integrating i get 62.5pi(27y-9y^2/2-y^3+y^4/4)

Then plugging in the points from -3 to 0 i finally end up with

-62.5pi(-81-81/2+27+81/4) =~14579 ft lbs

this is incorrect though.

I feel like i may have used a wrong substitution for r, but i do not know how i can relate r and y. can anyone help?

Thanks

You are saying height=(3-y). If y=(-3) at the bottom and y=0 at the top, does that really work? That would say height=6 and the bottom and height=3 at top. Doesn't seem right.

 

[doctordiddy]

You are saying height=(3-y). If y=(-3) at the bottom and y=0 at the top, does that really work? That would say height=6 and the bottom and height=3 at top. Doesn't seem right.

oh i see. I forgot to change that when i was trying to do it another way previously. Thanks for the help!