What percent of total tank capacity is being used?

[guitarra1]

Can you please help me to solve the following problem? Please show all steps...

A water storage tank has the shape of a cylinder with diameter 10 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 9 ft, what percentage, to the nearest one-tenth of a percent, of the total capacity is being used?

 

[soroban]

Re: calculus

Hello, guitarra1!

A water storage tank has the shape of a cylinder with diameter 10 ft.
It is mounted so that the circular cross-sections are vertical.
If the depth of the water is 9 ft, what percentage, to the nearest one-tenth of a percent,
of the total capacity is being used?

The cylinderical tank is oriented with its axis horizontal.
The circular ends have a radius of 5 feet.
The top one-foot segment is not filled with water.

Rotate the tank 90° and the end looks like this:

                |
              * * *
          *     |     *
        *       |     ::*
       *        |     :::*
                |     ::::
      *         |     ::::*
  - - * - - - - * - - +:-:* - - 
    -5*         |    4::::*5
                |     ::::
       *        |     :::*
        *       |     ::*
          *     |     *
              * * *
                |

Game Plan
Find the shaded area (empty space), and divide by the area of the circle, \(\displaystyle 25\pi\) ft².
. . That gives us the percentage of the tank that is not used.
Finally, subtract that value from one (100%).


The circle has the equation: \(\displaystyle \:x^2\,+\,y^2\:=\:25\;\;\Rightarrow\;\;y\:=\:\sqrt{25\,-\,x^2}\)

We want the area under the circle from \(\displaystyle x=4\) to \(\displaystyle x = 5\;\) ... times 2.

The shaded area is: \(\displaystyle \L\:A \;=\;2\int^{\;\;\;5}_4\sqrt{25\,-\,x^2}\,dx\)


Since you've been assigned this problem,
. . I assume you're familiar with Trig Substitution
. . and can complete the problem.