Changing order of integration double integral

[skate_nerd]

Homework Statement

Set up an integral for the ∫∫xydA for the region bounded by y=x, y=2x-2, and y=0. Set up the dxdy integral, then the dydx integral, then evaluate the simplest of the two.

Homework Equations

The Attempt at a Solution

I drew the region which was easy enough, and the easier integral to come up with was the dxdy integral with the y bounds from 0 to 2 and x from y to (y+2)/2. I'm pretty positive this one is correct, but my troubles show up when I try to come up with one for dydx. I know the x boundary is 0 to 2 but the y boundary is confusing me. I first tried from 2x-2 to x but that didn't give me the same answer as the dxdy integral. Then I tried a y boundary from 0 to x and that didn't work either. Any hints at what is giving me problems would be appreciated.

 

[MathematicalPhysicist]

Well, you have y between [x,2x-2] and x between [0,1];

Or x between [y,(y+2)/2] and y between [0,2], you look for the point of intersection of x and 2x-2 which occurs at x=2.

You might need to change the order of the limits, the picture you draw shoud supply you with enough information.

 

[voko]

What happens is that the lower limit y(x) has two parts in its domain, and its definition is different between them. Find the subdomains and the expressions.

 

[HallsofIvy]

Integrating in the order \int_{y=}\int_{x=} f(x,y) dxdy is straight forward. For \int_{x=}\int_{y=} f(x,y)dydx, the lower edge of the region is y= 0 until x= 1 where it becomes y= 2x- 2. So do that as the suim of two integrals.

 

[skate_nerd]

Ah I think I have it now. The order dydx is more complicated, and is a sum of two integrals. First with x from 0 to 1 and y from 0 to x and then the next x from 1 to 2 and y from 2x-2 to x. Evaluating that gives me the same answer as evaluating the single integral order dxdy that I already had.