Question

A tank in the shape of a right circular cone has height 12 feet and base radius 4 feet. The tank is inverted, with its vertex pointing down and base at the top. The tank contains a liquid with weight density 63 pounds per cubic foot, but is filled to a depth of 8 feet only. Set up, but DO NOT EVALUATE, an integral for the work (in foot-pounds) required to pump all the liquid present to a height one foot over the top of the tank.

Answer 1

V = int(π(y/3)^2, 0, 8)

(Definite integration of π(y/3)^2 with lower boundary 0 and upper boundary 8)

Explanation:

Set up cartesian axis (x and y) to the system.

Let y axis as the line of the centre of the cone, passing through its vertex and the centre of it's circular base. The x axis could be the 90 degree line to the y axis that passes the vertex. So the origin (0,0) is at the vertex.

I'm this setup , looking it as if we are looking it in 2 dimension, we'll see that there is a signature straight line on the x-y plane, which this line will form the cone as it revolute around the y-axis

Find the equation of the line:

Using height 12 and radius base 4, we can get the slope of the line

m = 12/4 = 3

It passes through origin, so the y-intercept is 0

Hence, y = 3x

Since the volume revolves around y-axis, we use the equation volume of revolution around y-axis

V = int(πx^2,a,b)

(Definite integration of πx^2 with lower boundary a and upper boundary b

Since y=3x

x = y/3

For this que

V = int(π(y/3)^2, 0, 8)

(Definite integration of π(y/3)^2 with lower boundary 0 and upper boundary 8)