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Math Help - Double Integrals over General Regions: Triangle

  1. #1
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    Double Integrals over General Regions: Triangle

    The problem is this:

    Evaluate the double integral. , D is the triangular region with vertices (0,0), (2,4), and (6,0).
    Solution Attempt:
    Line 1: y=2x
    Line 2: y=-x+6
    0=<x=<6

    Integrating ye^x dy dx yields 1/2y^2 e^x and when evaluated from 2x (bottom) to -x+6(top) it yields 1/2 e^x ((-x+6)^2 -4x^2); integrating this with respect to x yields e^x (-14+2x+x^2) and evaluating this from 0 to 6 yields -21-51e^6.

    This is a problem given online and says this is incorrect but i've done this by hand and in maple. Please point out what i've done wrong. Thank you.

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  2. #2
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    Quote Originally Posted by jdj0202 View Post
    The problem is this:

    Evaluate the double integral. , D is the triangular region with vertices (0,0), (2,4), and (6,0).
    Solution Attempt:
    Line 1: y=2x
    Line 2: y=-x+6
    0=<x=<6

    Integrating ye^x dy dx yields 1/2y^2 e^x and when evaluated from 2x (bottom) to -x+6(top) it yields 1/2 e^x ((-x+6)^2 -4x^2); integrating this with respect to x yields e^x (-14+2x+x^2) and evaluating this from 0 to 6 yields -21-51e^6.

    This is a problem given online and says this is incorrect but i've done this by hand and in maple. Please point out what i've done wrong. Thank you.

    When considering the y limits, draw a line that passes straight through the region from negative values of y to postive values. Clearly, that line ENTERS the region through the x axis, but which line does it EXIT the region through? Well there's one of two lines it exists through... it can exit through either  y = 2x or  y = x - 6 .

    Do the same with the x direction. Draw a line from negative values of x to positive values of x that passes straight through the region. You'll see that it passes through ONE line to enter the region ( y = 2x) and passes through ONE line to exit the region (  y = x - 6 ).

    That tells you that you should probably allow you x limits to be functions of y, and allow you y limits to be the consants. So:

     \frac{y}{2} \leq x \leq y + 6

     0 \leq y \leq 4

    So your integral should be:

     \int_0^4 \int_{\frac{y}{2}}^{y+6} ye^{x} \, dx \, dy
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