Finding Work Done with Integration

[armolinasf]

Homework Statement

There's a cone with height 12 feet and radius 4 feet. It is filled with water to a depth of 9 feet. Find the work required to pump the water over the top. The density of water is 62.4lbs/ft^3

work=force*distance & force=volume*density

The Attempt at a Solution

F=pi(r^2)*62.4 and r=(12-y)/3 by similar triangles

the distance should be 12-y

Putting this together, my integral looks like this:


62.4\pi/9 \int^{9}_{0}(12-y)^{3}dy

evaluating gives 37491 ft-lbs, but the answer in my book is 27788 ft-lbs

If someone can point out where I'm going wrong, it would be much appreciated.

 

[Dick]

If y is the distance from the bottom of the cone and r=(12-y)/3 then r=0 at the top of the cone and r=4 at the bottom. You've got the cone upside down.

 

[armolinasf]

Sorry I should have mentioned this, it specifies that the cone is actually pointing downwards

 

[SammyS]

Sorry I should have mentioned this, it specifies that the cone is actually pointing downwards

As Dick said: "If y is the distance from the bottom of the cone and r=(12-y)/3 then r=0 at the top of the cone and r=4 at the bottom."

That is pointing UP !

As you said, the cone is pointing downwards.

 

[armolinasf]

Wow I should have caught that...thanks for the help