reversing integration

[hils0005]

b]1. The problem statement, all variables and given/known data[/b]

Given Domain :\int\intf(x,y)dydx
0\leqx\leq1
x-1\leqy\leq2-2x
reiterate the integrals so the order is reversed

2. Relevant equations



3. The attempt at a solution
not really sure how to complete,
\int\intf(xy)dxdy
y+1\leqx\leq(2-y)/2
-1\leqy\leq2

Is this correct?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[HallsofIvy]

b]1. The problem statement, all variables and given/known data[/b]

Given Domain :\int\intf(x,y)dydx
0\leqx\leq1
x-1\leqy\leq2-2x
reiterate the integrals so the order is reversed

2. Relevant equations



3. The attempt at a solution
not really sure how to complete,
\int\intf(xy)dxdy
y+1\leqx\leq(2-y)/2
-1\leqy\leq2
It should be obvious that it is not correct. The first integral is a number. This integral will, after integrating, be a function of y. In order to be a number, the limits of integration on the "outside" integeral, with respect to y, must be numbers, not functions of y.

Draw a picture! In the original "outside" integral, x ranges from 0 to -1. For each x, y ranges from x-1 up to 2- 2x. Those are straight lines and you should see that the area is a triangle with vertices (0, -1), (2, 0) and (0, 2).

Now look at it from the side. To cover that entire area, y needs to vary from -1 to 1: those will be the limits on the "outer", dy, integral. For each y, x varies form 0 up to x= a function of y, given by the line making the right boundary. It looks to me like you will need to separate that into two integrals: y form -1 to 0 and then from 0 to 1.

[hils0005]

I have the verticies of the triangle at (2,0), (1,0), and (0,-1)?

wouldn't that mean dy would range from -1 to 2?

the lower part of the triangle dy from -1 to 0, x goes from 0 to y+1
ther upper part, dy ranges from 0 to 2, x goes from 0 to (y-2)/2

do you put an addition sign in between the the two integrals?

[HallsofIvy]

I have the verticies of the triangle at (2,0), (1,0), and (0,-1)?

wouldn't that mean dy would range from -1 to 2?

the lower part of the triangle dy from -1 to 0, x goes from 0 to y+1
ther upper part, dy ranges from 0 to 2, x goes from 0 to (y-2)/2

do you put an addition sign in between the the two integrals?
Then draw the graph more carefully. The two lines y= x- 1 and y= 2- 2x intersect at (1, 0). y= x-1 intersects x=0 at y= -1 and y= 2- 2x intersects x= 0 at y= 1. The vertices are, as I said, (0, -1), (0, 1), and (1, 0).

[gabbagabbahey]

...and y= 2- 2x intersects x= 0 at y= 1...

Really? :wink: