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Math Help - work pumping water into hemispherical tank

  1. #1
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    work pumping water into hemispherical tank

    A hemispherical tank of radius 6 feet is positioned so that its base is circular. a) How much work is required to fill the tank with water through a hole in the base if the water source is at the base?
    b) How much work is required to half fill the tank with water through a hole in the base if the water source is at the base?

    i got part a). i let x be the distance from the base to the top of the water layer and r be the radius of the water layer since each layer is a disc. so by pythagorean theorem, r = sqrt(36-x^2). so since the weight of water is 62.5 lbs/ft^3 the work done is integral (from 0 to 6) (62.5)(pi)(x)(36-x^2)dx and i got 20250pi ft lbs which was correct.

    however i can't seem to get my teachers answer for part b). i found the volume of the hemisphere which is 144pi and since we want to fill it to half, the volume of water will be 72pi. i then set up the equation 72pi = 144pi - (1/2)(4/3)(pi)(36-x^2)^(3/2). what i did was subtract the smaller hemisphere from the larger hemisphere since when you fill the half fill the tank the volume of the water won't be a hemisphere. i solved for x (the height of depth of water) and i got 3.649 ft. but my teacher's answer for the depth was 2.083 ft. how did she get that?
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  2. #2
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    I actually reproduced your answer, but I had to cut a hole in the middle of the water. No good. I think Moses was the last one known to have pulled that trick.

    I suggest you rethink your logic and try a different equation. It's a hemisphere and you are filling to 1/2. How can you get more than half the radius? Are you filling just the top part? That could be a gravity problem.

    Think about what you are saying. "Smaller hemisphere"? What is that? "Hemisphere" means something. Don't try to make it mean something else.

    There is a reason why you studied those volumes of solids of revolution. Here's your chance. You can do it the REALLY hard way, if you studied only shells:

    2\pi\int_{a}^{6}x\cdot \sqrt{36-x^{2}}\;dx\;+\;\pi a^{2} \sqrt{36-a^{2}}\;=\; 72\pi to get the appropriate x-value. The associated value of 'y' is what you need.

    Or, you can solve for it more directly if you are adept at the y-axis.

    \pi\int_{0}^{b} \left[\sqrt{36-y^{2}}\right]^{2}\;dy\;=\;72\pi

    Note: I'm not quite convinced on part 'a'. How does the water get to the top layer? I think tesseracting is out of the question. Are you sure it doesn't have to lift whatever water is already in there? Maybe it mixes instantaneously, or some other dubious assumption. It's an okay practice problem, I suppose, but I think I would not loan you my tractor to provide the hydraulics for this assignment.
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