A tank in the shape of a right circular cone is full of water. If the height of the tank is 15 feet and the radius of its top is 6 feet, find the work done

Question

A tank in the shape of a right circular cone is full of water. If the height of the tank is 15 feet and the radius of its top is 6 feet, find the work done in

(a) pumping the water over the top edge of the tank, ______ foot-pounds,

and (b) pumping the water to a height 15 feet above the top of the tank, _______ foot-pounds.

62.4 is the (weight) density of water in pounds per cubic foot.

Answer 1

This video is for the same problem: https://www.youtube.com/watch?v=B55EYe-fmbs

Answer 2

The idea here is to take an arbitrary sliver of the water and figure out how much work it takes to move that sliver. Here's a picture of a sliver.

https://i.stack.imgur.com/Heqek.png

The units in this problem also make it a little tricky. If you've taken physics you know the work needed to raise an object is mgh. If the mass has a density ρ and a volume V we can also write the work as ρVgh. Remember lbs is a measure of force and kilograms is a measure of mass. In these units the weight density is ρg.

We can call this ρ_w = ρg, and

ρ_w = 62.4 lbs/ft³

This makes the work of moving one sliver,

W = ρ_w V h

If we consider an arbitrary sliver a height h from the bottom of the cone its volume would be

dV = π r² dh

dV is the small volume of the sliver and dh is its height. As we have a right circular cone r=y and

dV = π y² dy

(a) We need to move the slab a height 15-y. This gives a work to move the sliver to the top of the cone as

dW = ρ_w π h² (15-y) dy.

For the total work you can integrate this expression from the bottom to the top of the cone.

W = ∫₀¹⁵ ρ_w π y² (15-y) dy.

This is just a standard integral at this point, there are no secret tricks here.

(b) In this case the height we need to move the sliver changes to 30-y. This gives the work as

W = ∫₀¹⁵ ρ_w π y² (30-y) dy.

Make sure to double check everything.