Solve the following word problem using integrals.

Question

A water storage tank has the shape of a cylinder with diameter 10 feet. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 8 feet, what percentage of the total capacity is being used? (Round your answer to one decimal place.)

I tried solving the question by using the formula (volume of water)/ (total volume of tank) and I used the integral ∫(from 0 to 2) √25-y^2 dy but my answer was not correct. Thank you in advance.

Answer 1

I'm assuming that by "depth of the water", the problem is referring to how high the water level is, thus there would be 8 feet of water, not 2 feet. Regardless, if there is 2 feet of water instead, then the final answer will be 100% minus the answer from this interpretation.

The way you tried to solve it is almost right, except that because we want the volume of a circular container rather than a half-circular container, we need to integrate 2*sqrt(25 - x^2) instead of just sqrt(25 - x^2).

Let's assume the length of the container is some number L.

To that end, the volume of the part of the container that is filled is given by ∫^x_-5 2L*√(25-t²) dt, for -5 ≤ x ≤ 5.

On the other hand, the volume of the entire container is given by 25πL

Therefore the ratio will give the fraction of the tank that is filled with water. Luckily, the L's will cancel when we simplify the ratio. Plugging in x = 3 to calculate a container filled with 8 feet of water yields

(2/25π) * ∫³_-5 √(25-t²) dt ≈ 0.857646... which is about 85.8%

Since it's asking for an approximation, I think it's safe to just input this equation into a graphing calculator and let the integration function do the work for you. If your question requires you to compute the integral by hand, you can use a trig substitution with x = 5sin(t) or x = 5cos(t). Doing this method gives you a result of

1/2 + (1/π)*arcsin(3/5) + 12/(25π) ≈ 0.8576215101... which is again about 85.8%

Answer 2

What you have is only part of the way towards the solution. If you solve your integral, you'll get the area of the cross-section of the unfilled part of the tank. Since the whole area is equal to 25π, if your area is represented by A, the answer would be (25π - A)/(25π).