Question

A tank on a water tower is a sphere of radius 30 feet. Determine the depths of the water when the tank is filled to one-fourth and three-fourths of its total capacity. (Note: Use the zero or root feature of a graphing utility after evaluating the definite integral. Round your answers to two decimal places.)

Answer 1

To determine the depths of water in the tank at one-fourth and three-fourths of its total capacity, calculate the height by multiplying the tank's radius by the cube root of the given fractions.

Step-by-step explanation:

To determine the depths of the water when the tank is filled to one-fourth and three-fourths of its total capacity, we need to find the heights of the water in the tank.

For one-fourth of the total capacity, the height of the water would be equal to the radius of the sphere multiplied by the cube root of (1/4). The cube root is used because the volume of a sphere is directly proportional to the cube of its radius.

Similarly, for three-fourths of the total capacity, the height of the water would be equal to the radius of the sphere multiplied by the cube root of (3/4).

Answer 2

Part 1) When the tank is 1/4 filled the depth of liquid is 19.581 feet.

Part 2) When the tank is 3/4 filled the depth of liquid is 40.42 feet.

Step-by-step explanation:

The volume of the sphere of radius R when it is filled up to a depth 'h' from the base equals

`V(h)=\int _(0)^(h)(\pi (2Rh-h^2))dh\\\\V(h)=\pi Rh^2-(\pi h^3)/(3)`

Case 1) When the tank is 1/4 full we have

`V=(1)/(4)* (4\pi R^3)/(3)`

Upon equating the values we get

`\pi Rh^2-(\pi h^3)/(3)=(1)/(4)* (4\pi R^3)/(3)\\\\Rh^2-(h^3)/(3)=(R^3)/(3)`

Putting R = 30 feet and solving for 'h' we get

h = 19.581 feet.

Case 2) When the tank is 3/4 full we have

`V_(3/4full)=V_(full)-V_(top1/4)`

`h_(3/4)=60-19.581=40.42feet`