Help a farmer out

[farmerboy127]

I am a farmer who loved Geometry while I was in High School, however that was years ago and I now have a question.

I have a tank that is in the shape of a cylinder which is lying down. If the tank has a diameter of 10 feet and a length of 23 feet, how many cubic feet are being used if the depth of the liquid inside the tank measures 2 feet.

If the tank was standing, this would be easy, but it's not. Is there a specific formula to figure a partially-filled cylinder?

 

[jacket81]

First, lets pretend we place the circular surface on an axis, with it moved (5-2)=3 feet horizonally down.
That gives you equation x^2+(y+3)^2=(10/5)^2.

Now we need to find the x-value at y=0, which gives you x^2+9=25,
so, x=squareroot(25-9) = 4.

So, to find square surface, you use (pi(10/2)^2*((2tan-1(4/3))/360)) -
(4*3) = 23.18238 - 12 = 11.18238. (Note:tan-1 means inverse tangent)

So, the cubic area is 11.18238*23.

I apologise for not having a picture.
I'm sure the equation would make more sence.
Maybe someone will post solution along with a picture.

 

[soroban]

Hello, farmerboy127!

I have a tank that is in the shape of a cylinder which is lying down.
If the tank has a diameter of 10 feet and a length of 23 feet,
how many cubic feet are being used if the depth of the liquid inside the tank measures 2 feet?

            * * *
          *       *
        *           *
       *             *

      *       O       *
      *       *       *
      *      /|\      *
            / |θ\ 5
       *   / 3|  \   *
        * /   |   \ *
        A *---+---* B
            * * *
              C

We want the area of segment $\displaystyle ABC$.

From a right triangle, we have: $\displaystyle \,\cos\theta\,=\,\frac{3}{5}\;\;\Rightarrow\;\;\theta\,=\,\cos^{-1}\left(\frac{3}{5}\right)\;\;\Rightarrow\;\;\theta\:\approx\:53.13^o\,\approx\:0.9273$ radians.

The area of sector $\displaystyle OACB$ is: $\displaystyle \,A_1\;=\;\frac{1}{2}r^2(2\theta)\;=\;\frac{1}{2}(5^2)(1.8546)\;=\;23.1825$

The area of triangle $\displaystyle OAB$ is: $\displaystyle \,A_2\;=\;\frac{1}{2}(8)(3)\;=\;12$
$\displaystyle \;\;$(Note that $\displaystyle AB\,=\,8$)


Therefore, the area of the segment is: $\displaystyle \,A_1\,-\,A_2\;=\;23.1825\,-\,12\;=\;11.1825\;\text{ft}^2$.

Multiply by the length (23 feet) to get the volume of the liquid.