# Changing order of integration double integral

1. Mar 18, 2013

### skate_nerd

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

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.

2. Relevant equations

3. 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.

2. Mar 18, 2013

### 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.

3. Mar 18, 2013

### 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.

4. Mar 18, 2013

### HallsofIvy

Staff Emeritus
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.

5. Mar 18, 2013

### 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.