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

Answer:

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

$V(h)=\int _{0}^{h}(\pi (2Rh-h^2))dh\\\\V(h)=\pi Rh^2-\frac{\pi h^3}{3}$

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

$V=\frac{1}{4}\times \frac{4\pi R^3}{3}$

Upon equating the values we get

$\pi Rh^2-\frac{\pi h^3}{3}=\frac{1}{4}\times \frac{4\pi R^3}{3}\\\\Rh^2-\frac{h^3}{3}=\frac{R^3}{3}$

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

$V_{3/4full}=V_{full}-V_{top1/4}$

$h_{3/4}=60-19.581=40.42feet$