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Math Help - multiple integral

  1. #1
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    multiple integral

    I have a question about evaluating the iterated integral

    its a homework problem i was doing ok with the double integrals but the triple confuse me


    int(int(int(x*cosy,x,y,z,0,4,0,pi/2,0,1-x)))

    I think that is how you right it well i know it the problem but i was trying to get the signs to actually come up
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  2. #2
    Super Member Aryth's Avatar
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    It should look like this:

    \int_0^4\int_0^{\frac{\pi}{2}}\int_0^{1-x}xcosy~dx~dy~dz

    Verify that this is the right integral and I'll start working on it.
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  3. #3
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    Thank you !!!!
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  4. #4
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    yes this is correct!!!
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  5. #5
    Super Member Aryth's Avatar
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    Well, here's my work so far:

    I have:

    \int_0^{1-x} xcosy~dx = \frac{1}{2}x^2cos(y)]_0^{1-x}

    =\frac{1}{2}(1-x)^2cos(y)

    And then:

    \int_0^{\frac{\pi}{2}} \frac{1}{2}(1-x)^2cos(y)~dy

    =\frac{1}{2}(x-1)^2sin(y)]_0^{\frac{\pi}{2}}

    = \frac{1}{2}(x-1)^2

    And then:

    \int_0^4 \frac{1}{2}(x-1)^2~dz

    =\frac{z}{2}(x-1)^2]_0^4

    =\frac{4}{2}(x-1)^2 = 2(x-1)^2
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  6. #6
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    Dont understand

    But i dont get it because some how the answer is supposed to be -40/3
    and if you have already evaluated it from o to 4 how is that possible
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  7. #7
    Behold, the power of SARDINES!
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    \int_0^4\int_0^{\frac{\pi}{2}}\int_0^{1-x}xcosy~dz~dy~dx

    I think the order of integration got messed up
    we need to do z first

    \int_0^4\int_0^{\frac{\pi}{2}}zxcos|_{0}^{1-x}~dy~dx

    \int_0^4\int_0^{\frac{\pi}{2}}(x-x^2)cos(y)~dy~dx

    \int_0^4(x-x^2)sin(y)|_{0}^{\pi/2}~dx

    \int_{0}^{4}x-x^2~dx=\frac{1}{2}x^2-\frac{1}{3}x^3|_{0}^{4}=8-\frac{64}{3}=-\frac{40}{3}
    Last edited by TheEmptySet; May 6th 2008 at 10:18 PM.
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