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Thread: Tough Applied Integration Question

  1. #1
    May 2008

    Tough Applied Integration Question

    Hi there folks,
    Been browsing your forums and figured you were the best to maybe help me out. Been having quite a bit of trouble with the following:

    A trough is 1m long and semi-circular in cross-section with a radius of 100mm. The trough is initially horizontal and is filled with water. One end of the trough is lifted such that the angle between it and the horizontal is x degrees. Find a relationship/s between the volume of water remaining in the trough and the angle x.

    I know that the key to this problem is how the axes are defined. Unfortunately thats all i can figure.. Any help would be very much appreciated.
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  2. #2
    Eater of Worlds
    galactus's Avatar
    Jul 2006
    Chaneysville, PA
    I will give this a go using a triple integral. I will give my 2 cents. If I have an epiphany, I will certainly be back.

    What we have is half a cylinder which creates the semi circle.

    Keep the tank horizontal and look at the water surface as a tilted plane. Take

    the tank as the cylinder $\displaystyle x^{2}+y^{2}=r^{2}$, which goes

    along the z-axis. I suppose it doesn't matter which axis you use though.

    Then, calculate the volume by integrating with respect to z, with the sections

    perpendicular to the z-axis. The sections are segments of a circle.

    I would think it would depend on two different scenarios. One with the top

    end(the end raised in the air) being partially covered and then the case when

    the water is down enough that the top end is dry.

    Here is an integral I came up with. Admittedly, I am not 100% comfortable. See what you think.

    $\displaystyle \int_{0}^{r}\int_{0}^{\sqrt{r^{2}-y^{2}}}\int_{0}^{rcot({\theta})}dzdxdy=\frac{{\pi} r^{3}}{4}tan({\theta})$

    I will look again later. I have a feeling we may need another integral for the other case. If you find anything, let me know. Cool problem.
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