# Thread: changing the order of integrals

1. ## 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

2. Originally Posted by James0502
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:

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

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

3. Originally Posted by James0502
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:
$\displaystyle \int_{x=0}^a\int_{y= 0}^x sin(x+y)dydx$
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:
$\displaystyle \int_{y=0}^a\int_{x=y}^1 sin(x+y)dxdy$

(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.)