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Math Help - integration; work; missing 2

  1. #1
    Junior Member Tclack's Avatar
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    integration; work; missing 2

    A vat (shown in attachment) contains water 2 m deep. Find work required to pump all water out of top of vat. (weight density of water = \rho= 9810 N/m^3)

    W=\int^b_a F(x)dx

    weight of water= F(x)=V\rho

    put my figure on the coordinate axis so I came up with general equation for the base, with respect to height (using a y=mx+b format):


    h=\frac{3}{2}b-3
    but, I want to go down a positive depth, so I negated the equation:
    h=-\frac{3}{2}b+3
    b= (h-3)\frac{-2}{3}
    b=\frac{2}{3}(3-h)---------------------------------(1)

    As you can see, the above equation follows correctly, at a depth(h) of 0, we have a base of 2, at a depth(h) of 3 we have a base of 0, so the equation is valid, so far no mess-ups... I hope

    Now, the general equation for a triangular volume is: Volume= V=\frac{1}{2}bhL (b=base, h=height, L=length)
    But, because my equation for b is only taking one half of the triangle, I double it:
    V=2(\frac{1}{2}bhL)=bhL-------------------------------(2)
    substituting (1) into (2) I get
    V=\frac{2}{3}(3-h) hL= L=6m so
    V=4(3h-h^2)

    therefore
    W=\int^3_1 F(x)dx=\int^3_1 V\rho dx=\int^3_1 4(3h-h^2)(9810) dx
    W=4(9810)(\frac{3h^2}{2}-\frac{h^3}{3})\mid^3_1
    W=130800
    ...which is exactly half of the answer, I just can't seem to find that missing 2 despite being so concise. I remembered to double the volume of my second figure to take into account the full triangle base. Please help me find this... it's so close
    Attached Thumbnails Attached Thumbnails integration; work; missing 2-work.jpg  
    Last edited by Tclack; April 24th 2010 at 02:20 PM. Reason: Add figure
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  2. #2
    Junior Member Tclack's Avatar
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    I found this video
    it's very similar. That guy made little rectangular boards. I tried it...it worked. What's wrong with what I did?
    Last edited by Tclack; April 24th 2010 at 08:23 PM. Reason: error
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  3. #3
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    You seem to be looking at things rather oddly.

    1) Why do you care about the area of a triangle? The ONLY thing the triangular shape does for you is establish the relationship between the depth and the width of the layer being pumped.

    2) Do it one piece at a time and don't guess. You are going to pump one layer at a time. Do NOT try to do the whole triangle at once.

    3) The cross section of each layer is a rectangle, not a triangle.

    Depth / Height to Pump. Set this simply, as you have done, as 'h' for h between 1 and 3. This is not the only way to do it. ANY function that goes from 1 to 3 while the limits of integration run will be sufficient.

    This gives: \int_{1}^{3}h\cdot (Width.of.Layer) \cdot 6 \cdot 9180\;dt

    Width of Layer: You noticed that this is related to water depth, and not height to pump. That was excellent. Simply (4/3)(3-h).

    This gives: \int_{1}^{3}h\cdot (4/3)(3-h) \cdot 6 \cdot 9180\;dt

    4) Please notice how in this final form all the components are transparent. Rather than some algebraiclly simplified version that makes no sense, you still can see where each component contributes to the final result.
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  4. #4
    Junior Member Tclack's Avatar
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    I see what you mean. I am taking a rectangular cross section, not a triangular...

    My Error: I guess I just got hung up that my answer was just half of the actual answer, so I focused to much at looking for that error in the algebra rather than the formula itself.

    My half-answer must be a coincidence then, right?
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  5. #5
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    You were also hung up in the algebra. You should be able to write these down without a lot of fuss. There are only four components:

    1) Distance
    2) Cross-section
    3) \rho
    4) The previous three need to relate to your limits of integration.

    Focus on one piece at a time. Also, keep in mind that it is quite likely there is more than one method - of course, only one unique solution.
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