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Math Help - Doubble Integral

  1. #1
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    Doubble Integral

    double integral of (x^2*tanx+y^3+4)dR. Over the region x^2+y^2(greater than or equal to)2

    our teacher instructed us not to change the variables into polar coordinates, so I sticked to using the Cartesian coordinate system. Limits of integration since circle is symmetric to both axis it could be interchanged, I made my limits as
    -(2)^(1/2)<x<(2)^(1/2),
    -(2-x^2)^(1/2)<y<(2-x^2)^(1/2)..


    So as I said I separately integrated the variables, with "y^3" going to zero and "4" turning into 8pi.
    But my problem is about the double integral of x^2*tanxdR. At first , I integrated it with respect to y,
    so the result is Integral of (2)((2-x^2)^(1/2)*x^2*tanx) dx.
    But at this point, I don't know what to do. I couldn't figure out how to take the anti derivative , even Wolfram could not give me the anti derivative. But it could give me the answer to this definite integral that is 0. Not allowed to use taylor series either,
    Thanks in advance for your help.
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  2. #2
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    Re: Doubble Integral

    I have no idea what you mean by '"y^3" going to 0' or '"4" turning into 8pi' The anti-derivative of y^3 is, of course, y^4/4 but how does it "go to 0"? The anti-derivative of 4 (with respect to y) is 4y. How does that "turn into 8pi"? I also have a problem with "Over the region x^2+ y^2\ge 2". That is an unbounded region. You will certainly NOT give a finite result. Did you mean to say "less than or equal to 2"?
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  3. #3
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    Re: Doubble Integral

    First the region is x^2+y^2<(or equal )2 sorry should have made it in symbols.
    second, y^3 going to zero is the result of its definite integral y^4/4 with respect to y. lower bound y=-(2-x^2)^(1/2) to y=(2-x^2)^(1/2) so 1/4[(2-x^2)^2-(2-x^2)^2]
    and that same goes to 4 since double integral of a constant is 4*(area of the region) right? so 4*(2pi)=8pi
    My initial question is about the double integral of (x^2*tanx)dR.
    with the limits of :
    -(2)^(1/2)<x<(2)^(1/2),
    -(2-x^2)^(1/2)<y<(2-x^2)^(1/2). I first took the integral with respect to y resulting to another integrand that is (2)((2-x^2)^(1/2)*x^2*tanx).
    so integral of [(2)((2-x^2)^(1/2)*x^2*tanx] dx with x=-(2)^(1/2) to x=(2)^(1/2).

    wolfram tells me it is zero(Wolfram|Alpha Widgets: "Definite Integral Calculator" - Free Mathematics Widget). so the answer to the double integral must be 8pi.

    question. How did it become zero?
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