Geometry: cylinder, r = 2, L = 6, h = 3: find depth if vert.

[jdigrazia]

I am trying to help my son with a math problem. Here it is:

A cylindrical water tank with a radius 2 feet and length 6 feet is filled with water to a depth of 3 feet when in the horizontal position. If the tank is turned upright, what is the depth of the water? Give answer in terms of pi.

I understand how to calculate the volume of the whole tank but I cannot remember how to calculate a partially filled tank.

Help and guidance would be appreciated.

Thanks
Joe

 

[soroban]

Re: Geometry

Hello, Joe!

A cylindrical water tank with a radius 2 feet and length 6 feet
is filled with water to a depth of 3 feet when in the horizontal position.
If the tank is turned upright, what is the depth of the water?
Give answer in terms of \(\displaystyle \pi.\)

To find the volume of water in the tank when it is horizontal
. . takes a bit of extra thinking and work.
[I'll assume we can't use Calculus.]

Look at the end of cylindrical tank.

              * * *
          *     |D    *
      A * - - - + - - - * B
       *  *     |     *  *
            *  1|   * 2
      *       * | *       *
      *         *         *
      *         |O        *
                |
       *        |        *
        *       |2      *
          *     |     *
              * * *

\(\displaystyle \text{The circle has center }O\text{, radius 2: }\:OA = OB = 2\)

\(\displaystyle \text{It is filled to a height of 3 ft: }\:OD = 1\)

\(\displaystyle \text{We have right triangle }ODB\text{, with side }OD = 1\text{ and hypotenuse }OB = 2.\)
\(\displaystyle \text{This is a 30-60 right triangle: }\B = \sqrt{3},\;\angle BOD = 60^o\)
. . \(\displaystyle \text{Hence: }\:AB = 2\sqrt{3},\;\angle AOB = 120^o\)

\(\displaystyle \text{The area of the circle is: }\:A \:=\:\pi r^2 \:=\:\pi(2^2) \:=\:4\pi\text{ ft}^2\)

\(\displaystyle \text{Since }\angle AOB = 120^o\text{, the area of the sector }AOB\text{ is }\tfrac{1}{3}\text{ of the circle.}\)
. . \(\displaystyle A_{\text{sector}} \:=\:\tfrac{1}{3}(4\pi) \;=\;\frac{4\pi}{3}\text{ ft}^2\)

\(\displaystyle \text{The area of }\Delta AOB\text{ is: }\:\tfrac{1}{2}(\text{base})(\text{height}) \:=\:\tfrac{1}{2}(2\sqrt{3})(1)\)
. . \(\displaystyle A_{\text{triangle}} \:=\:\sqrt{3}\text{ ft}^2\)
\(\displaystyle \text{The area of the segment is: }\:\frac{4\pi}{3} - \sqrt{3} \;=\;\frac{4\pi - 3\sqrt{3}}{3}\text{ ft}^2\)
. . This is the area that is NOT covered with water.

\(\displaystyle \text{Then: }\:A_{\text{water}} \;=\;A_{\text{circle}} - A_{\text{segment}} \;=\;4\pi - \frac{4\pi -3\sqrt{3}}{3} \;=\;\frac{8\pi + 3\sqrt{3}}{3}\)

\(\displaystyle \text{Hence: }V \;=\;\text{(Area)} \times \text{(length)} \;=\;\frac{8\pi+3\sqrt{3}}{3}\times 6 \;=\;2(8\pi+3\sqrt{3})\)


\(\displaystyle \text{The volume of the upright cylinder is: }\:V \:=\:\pi r^2h\)

\(\displaystyle \text{We have: }\:r = 2,\:V \,=\,2(8\pi + 3\sqrt{3})\)

\(\displaystyle \text{And we have: }\:2(8\pi + 3\sqrt{3}) \;=\;\pi(2^2)h\)

. . \(\displaystyle \text{Therefore: }\:h \:=\:\frac{8+3\sqrt{3}}{2\pi}\)

 

[jdigrazia]

Hi Soroban

Thanks for the help. I understood exactly how you completed it (it brought back math memories from days long past) and my son also understood it and explain it to me which is really the point.

I need some clarification at the end when you calculated the Area of the water. I understand that you multiplied the Area of the circle by 3/3 to get common denominator. But how did you get
8(pi) plus 3(sq root of 3) it looks like it should be Minus not Plus?

Still a little confused.

Joe

 

[Denis]

Ya Joe, but only "looks" confusing :wink:

We have 4p - (4p - 3sqrt(3)) / 3 ; that's:
12p/3 - (4p - 3sqrt(3))/3 ; same as:
[12p - (4p - 3sqrt(3))] /3 ; removing brackets:
[12p - 4p + 3sqrt(3)] /3 ; wrapping up:
[8p + 3sqrt(3)] / 3 ; BINGO!