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Math Help - please check integral problem

  1. #1
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    please check integral problem

    basin has a form of cylinder with height of 0,5 meters and background radius of 0,3 meters. calculate a work that is needed to evacuate the water from the basin.

    my solution: area of background = Pi \cdot R^2
    as water has density of 1000 and F=m \cdot g then
    A=m \cdot g \cdot x=882 \cdot Pi \cdot x and integrating the last expression from 0 to 0,5 i get 346,36 but an answer given for this problem is 346,54.

    basin has a form of half of a sphere with radius 0,1 meters. calculate a work needed to evacuate water from it.

    my solution: are of circle that is x meters under the center is
    Pi \cdot r^2 = Pi(R^2 - x^2) = Pi(0,01 - x^2)
    A=m \cdot g \cdot h =Pi(0,01 - x^2) \cdot 10^4 \cdot x
    and as i integrate it from 0 to 0,1 i get 0,79, but the answer given is 1,54.
    please help me find my mistakes
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  2. #2
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    For the first one, I did the example myself and used g = 9.8 and got your value. Using a value of g = 9.805, I got 346.537, which rounded gives the desired result. However, I looked it up and got a value of 9.806 from Gravity of Earth - Wikipedia, the free encyclopedia (third paragraph).

    If you take 9.789 at the equator, you get 345.97
    And if you take 9.832 at the poles, you get 347.492

    So I guess you should expect a small variation in your answer, depending on the exact constants used.

    For the second part, it is somewhat ambiguous. Which way is the hemisphere facing?
    Last edited by Vlasev; July 29th 2010 at 04:14 AM.
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  3. #3
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    ok, thanks for the first one. it is good if this variation in answer is caused by constants. the main thing i wanted to check whether i have applied the integral correctly.
    for the second i thought the same way, but only difference would be that area of background is function of x.
    Attached Thumbnails Attached Thumbnails please check integral problem-img_1329.jpg  
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  4. #4
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    No problem. I think you have applied the integral correctly. For the second case, judging by your picture, I think you've done the integration correctly. I did the calculation too and again used 9.805. I got 0.770083, which is 1/2 of 1.540166, which rounded to 3 sig figs is 1.54. It's possible that the answer in the book is off by a factor of 2 itself!
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  5. #5
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    thanks alot. hope that it is a wrong answer in the book
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