A cone-shaped water reservoir is 20 ft. in diameter and 20 ft. deep. If the reservoir is filled to a depth of 10ft., then write the integral which represents the amount of work required to pump all the water to the TOP of the reservoir.

The work required is proportional to the height, measured from the bottom.

This integral is S (pi*R(H)^2)*(density)*g H dH. S denotes the integral sign

The cone radius as a function of height is R(H) = H/2

Integrate from H = 0 to h = 20 ft

In these British units, (density* g = 62.4 lb/ft^3. The answer will be in ft-lb

To find the integral that represents the amount of work required to pump all the water to the top of the reservoir, we need to calculate the work done in pumping each slice of water and then sum up all these slices.

First, let's consider a thin slice of water at a certain depth inside the cone-shaped reservoir. The width of this slice can be determined by similar triangles. Since the diameter of the reservoir is 20 ft, the width of the slice at a certain depth will be (20/20) times the depth, or simply equal to the depth itself.

Next, let's determine the volume of this thin slice of water. The volume of a cone is given by the formula (1/3) * π * r^2 * h, where r is the radius of the cone's cross-section and h is the height of the cone. In our case, the radius is half the diameter, so it's 10 ft. The height of the slice is the width, which we determined to be the depth.

Now, let's find the mass of this thin slice of water. The density of water is constant, so we can assume it as ρ. The mass of the slice can be calculated as the volume times the density.

Finally, the work done in pumping this thin slice of water is equal to the product of the height of the slice (the depth) and the weight of the slice (the mass times the acceleration due to gravity, g).

Therefore, the work done in pumping this thin slice is given by:
work = depth * (ρ * (1/3) * π * r^2 * depth) * g

To get the total work required to pump all the water to the top of the reservoir, we need to integrate this expression over the range of depths from 0 to 10 ft. Let's use the variable "x" to represent the depth.

The integral that represents the amount of work required is:
∫[0 to 10] x * (ρ * (1/3) * π * (10/20)^2 * x) * g dx

Simplifying this integral will give you the final answer.