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Math Help - Evaluating integral using change of variables

  1. #1
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    Evaluating integral using change of variables

    Hey guys, just having a bit of trouble getting started with this.

    Have to use a change of variables to evaluate the integral
    \int\int_{D}sin(x+2y)sin(3x-y)dxdy
    where D={ (x,y)\in\Re^2 | 0 \leq x+2y \leq (\Pi/2) and 0\leq 3x-y\leq (\Pi/2) }

    I assume we use polar coordinates, I'm just a bit lost with the inequalities in D
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  2. #2
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    Quote Originally Posted by MrSplashypants1 View Post
    Hey guys, just having a bit of trouble getting started with this.

    Have to use a change of variables to evaluate the integral
    \int\int_{D}sin(x+2y)sin(3x-y)dxdy
    where D={ (x,y)\in\Re^2 | 0 \leq x+2y \leq (\Pi/2) and 0\leq 3x-y\leq (\Pi/2) }

    I assume we use polar coordinates, I'm just a bit lost with the inequalities in D
    substitution is what you're supposed to do here. let x+2y=u, \ 3x - y= v, find the Jacobian of this transformation, etc. you'll end up with a very simple integral.

    by the way that's \frac{\pi}{2} not \frac{\Pi}{2}.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    substitution is what you're supposed to do here. let x+2y=u, \ 3x - y= v, find the Jacobian of this transformation, etc. you'll end up with a very simple integral.

    by the way that's \frac{\pi}{2} not \frac{\Pi}{2}.
    Alright, so we define
    g: D^* \longrightarrow D
    (u,v) \longmapsto (x,y)
    where u=x+2y and v=3x-y
    Then D^* is the square region bounded by 0 \leq u \leq (\pi/2) and 0\leq v \leq (\pi/2)

    And so the original integral becomes \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}sin(u)sin(  v)|detJ_{g} (u,v)|dudv

    where the determinant =  \frac{1}{7}

    and so the integral comes out as just  \frac{1}{7} as well?
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  4. #4
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    Quote Originally Posted by MrSplashypants1 View Post
    Alright, so we define
    g: D^* \longrightarrow D
    (u,v) \longmapsto (x,y)
    where u=x+2y and v=3x-y
    Then D^* is the square region bounded by 0 \leq u \leq (\pi/2) and 0\leq v \leq (\pi/2)

    And so the original integral becomes \int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}sin(u)sin(  v)|detJ_{g} (u,v)|dudv

    where the determinant =  \frac{1}{7}

    and so the integral comes out as just  \frac{1}{7} as well?
    yes. \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \sin u \sin v |\det J| \ du \ dv = \frac{1}{7} \left( \int_0^{\frac{\pi}{2}} \sin u \ du \right)^2=\frac{1}{7}.
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  5. #5
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    Thank you very much sir
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