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Math Help - cylinder with water

  1. #1
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    cylinder with water

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

    cylinder with water-spheres.gif


    Find the volume and total area of a cube with edge 2k.

    A right triangular prism has height 20 and base edges 5, 12, and 13. Find the total area.
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  2. #2
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    Why would someone spell \pi as pie?. You can't really think it's spelled that way?.
    It's not a pastry, it's a letter of the Greek alphabet. PI


    There's different methods to find the solution to #1. Calculus or non-calculus.

    The height of the empty region in the tank is given by 1=2(1-cos(\frac{\theta}{2}))

    Solving for theta, we find \theta=2cos^{-1}(\frac{1}{2})=\frac{2\pi}{3}

    The volume of a circular segment is given by \frac{1}{2}r^{2}(\theta-sin{\theta})=2(\theta-sin{\theta})=2(\frac{2\pi}{3}-sin(\frac{2\pi}{3}))=\frac{4\pi-3\sqrt{3}}{3}\approx{2.456....}

    The length of the tank is 6, so the volume of this portion of the tank is

    8\pi-6\sqrt{3}

    Subtract this from the volume of the tank which is 24\pi

    and get 24\pi-(8\pi-6\sqrt{3})\approx{60.66}

    Then use the volume of a cylinder formula and solve for h:

    4{\pi}h=24\pi-(8\pi-6\sqrt{3})

    h\approx{4.83}

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

    Calc method:

    12\int_{-2}^{1}\sqrt{4-x^{2}}dx\approx{60.66...}

    Now, the same method as before and we get the same answer.
    Last edited by galactus; July 31st 2007 at 01:33 PM.
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  3. #3
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    Whoops, sorry... wasn't thinking. I have habit of that--I've spelled 'edition' as addition, etc...
    Thank you!
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  4. #4
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    Since I'm supposed to put my answer in terms of pi, can I simplify this statement further? Could I just divide everything by two?

    <br />
4{\pi}h=24\pi-(8\pi-6\sqrt{3})<br />

    Oh, and does anybody want to try the other two problems?
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  5. #5
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    Quote Originally Posted by blindspot View Post
    ...
    Find the volume and total area of a cube with edge 2k.

    A right triangular prism has height 20 and base edges 5, 12, and 13. Find the total area.
    Hello,

    as you may know the volume of a cube is calculated by:
    V_{cube}=a^3. (a = length of edge). Plug in the given value and you'll get V = 8k

    The surface area of a cube is
    A_{cube}=6 \cdot a^2. Plug in the given value and you'll get A = 24k
    = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

    The base triangle is a right triangle because 5 + 12 = 13. Thus the base area and the top area form a rectangle with length = 12 and width = 5.

    The surface area of a right prism is calculated by:

    A_{prism}=base + top + perimeter_{base} \cdot height. Plug in the given value and you'll get:

    A_{prism}=5 \cdot 12 + 30 \cdot 20 = 660
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  6. #6
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    Quote Originally Posted by blindspot View Post

    Oh, and does anybody want to try the other two problems?
    If you have other problems, feel free to post them in new threads.

    -Dan
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