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Math Help - changing the order of integrals

  1. #1
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    changing the order of integrals

    evaluate (double integral) sin(x+y)dxdy

    over the region R bounded by the line y = x, y = 0 and x = a where a > 0

    show that the result is the same of the order is reversed

    My first thought was the integrate wrt y from x to 0 then wrt x from a to 0
    but then how do i reverse the integral?

    many thanks
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  2. #2
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    Quote Originally Posted by James0502 View Post
    evaluate (double integral) sin(x+y)dxdy

    over the region R bounded by the line y = x, y = 0 and x = a where a > 0

    show that the result is the same of the order is reversed

    My first thought was the integrate wrt y from x to 0 then wrt x from a to 0
    but then how do i reverse the integral?

    many thanks
    Draw the region of integration.

    Then you should see that:


    I = \int_{x=0}^{x=a} \int_{y=0}^{y =x} \sin (x + y) \, dy \, dx


    I = \int_{0=0}^{y=a} \int_{x=y}^{x=a} \sin (x + y) \, dx \, dy
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  3. #3
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    Quote Originally Posted by James0502 View Post
    evaluate (double integral) sin(x+y)dxdy

    over the region R bounded by the line y = x, y = 0 and x = a where a > 0

    show that the result is the same of the order is reversed

    My first thought was the integrate wrt y from x to 0 then wrt x from a to 0
    but then how do i reverse the integral?

    many thanks
    The region bounded by y= x, y= 0, and x= a, for a> 0, is a triangle with vertices at (0,0), (a, 0) and (a, a). The limits of integration of the "outer" integral must be a constant so you pick the smallest and largest values of the variable in that region. If we choose to do the "outer" integral with respexct to x, then x can be as small a 0 or as large as a so we use 0 and a as limits. For each x (draw a vertical line on the graph to represent that), y can vary from 0 up to the line y= x. The limits on the inner integral are 0 to x:
    \int_{x=0}^a\int_{y= 0}^x sin(x+y)dydx
    Notice that y goes from 0 up to x, not "from x to 0" as you suggest. x> 0 for any point in this region.

    To reverse the order, what are the smallest and largest values of y in that region? They are, of course, 0 and a again. Now draw a horizontal line representing a fixed y. In order to cover the same triangle, its left endpoint is on the line y= x and its right endpoint is on the line x= a. x must go from y up to 1: the limits on x are y and 1:
    \int_{y=0}^a\int_{x=y}^1 sin(x+y)dxdy

    (Notice the "x= " and "y= " on my lower limits. I recommend that. It's easy to get confused as to whichy limits are which and that helps keep them straight.)
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