Can anyone help me figure out this Calculus II question?

Question

This question was on my test and the answer was 3,272,500.

A tank is in the shape of a right circular cone of altitude 25 feet and base radius 20 feet with its vertex at the ground. (Think of an ice cream cone with its point facing down.) The tank is externally braced to prevent its falling over. If the tank is full of a fluid which weighs 50 pounds per cubic foot, find the work done, in foot pounds, correct to the nearest hundred footpounds, in pumping all of the fluid out of the top of the tank.

I know that the Volume of a cone= 1/3(pie)r^2h
I tried working the problem in the same manner as the spherical cylinder in the book except I used 1/3(pie) as opposed to just pie. I think the problem is my calculations of the radius being incorrect. The definite integral I came up with was

25
Int (1/3)(50)(pie)x(40-x)^2
0
as my r^2 of the ith interval was (40ci -ci^2)

Again help is greatly appreciated.

Okay the first answer worked but I don't know how he got the 20000. I multiplied radius of 20^2 times the height of 25 and I get 10000. What did I miss.

Answer 1

First, you need the volume. The formula yields: V = pi rh/3 = 20000pi/3 cubic feet. Since it weighs 50 pounds per cubic foot, the ratio 50 lbs/ft.^3 = 1. So multiplying this to our volume equation, V = 20000pi/3 ft.^3 = 20000pi/3 ft.^3 * 50lbs/ft.^3 = 100000pi/3 lbs. The work function is (25-x)100000pi/3 = W(x). Integrate this result from x=0 to x=25, and you get the answer.

Answer 2

dW=dF p (where dF is the change in Force and p is the distance)

Force is equal to the amount of mass times gravity. With a uniform density, the mass is proportional to the volume. What you have is an increasing amount of mass as you move up the cone. The rate of change in the Force is proportional to the RATE OF CHANGE in the volume. In other words, you need dV instead V.

dV = PI r^2 dh

The radius is 4/5 the height, so you can rewrite your equation as:

dV = PI (4/5 h)^2 dh = PI 16/25 h^2 dh

The change in Force, dF, is equal to 50 lb/ft^3 * dV
dF=50*dV

The distance moved (p) will be equal to 25-h (where h is whatever height you happen to be at)

You start with: dW = dF p and substitute:

dW= 50* (16/25 PI h^2) (25-h) dh

The rest is just some simple algebra and integration.

You should eventually get W = 8/3 PI 25^4 , which does equal about 3272492

Answer 3

Divide the Cone into thin cylindersstacked one on top of each other.

Now, let the height of a cylinder be dx, derive its radius in terms of x and the angle of the cone.

Use the formula for work

dW= PdV

where dV is the volume of the cylinder.

Integrate it from 0 to h, (height of cone)