# Double integral problem help appreciated!

1. Nov 15, 2007

### engineer_dave

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

Evaluate the integral shown ( I have the file with the given integral attached here).

2. Relevant equations

3. The attempt at a solution

So what i did was change dy dx into dx dy. Then i integrated y so the whole thing becomes 2x - y^3. I plugged the values (1+x) and (1-x) into y^3. After here, I got a really complicated thing which probably means im on the wrong track. The answer given was -13/6. Can you please help? Thanks!

#### Attached Files:

• ###### Double integral.JPG
File size:
5.2 KB
Views:
70
2. Nov 15, 2007

### HallsofIvy

Staff Emeritus
I'm not at all clear what you mean by "change dydx into dxdy". I thought at first you meant that you changed the order of integration so that you would be integrating with respect to x first but then you say "Then i integrated y" which I take to mean you integrated 2x- 3y2 with respect to y. The result of that is NOT, however, 2x- y3, it is 2xy- y3. Substitute y= 1+x and y= 1-x into THAT and subtract.

Last edited: Nov 15, 2007
3. Nov 15, 2007

### engineer_dave

yea i want to change the order of integration so it becomes dx dy. Then what do I have to do?

4. Nov 15, 2007

### HallsofIvy

Staff Emeritus
You had better have a really good reason for wanting to do that! It's much messier than integrating the way you have it.

The line y= 1+ x forms the upper boundary of the region and the line y= 1-x forms the lower boundary. The vertical line x= 1 is the right boundary. (y= 1+x and y= 1-x cross at (0,1) so the two lines form the "left boundary".) The region is a triangle with vertices at (0,1), (1,2) and (1, 0).

In order to cover that region y has to go from 0 up to 2. HOWEVER the left boundary involves two different lines and so two different formulas. For y between 0 and 1, the left boundary is given by the line y= 1- x or x= 1- y. x must go from 1-y to 1 for each y. For y between 1 and 2, the left boundary is given by the line y= 1+ x or x= y-1. x must go from y-1 to 1 for each y. That means you will have to do the integral in two different parts:
[tex]\int_{y= 0}^1\int_{x= 1-y}^1 (2x- 3y^2)dx dy+ \int_{y= 1}^2\int_{x= y-1}^1(2x-3y^2)dx dy[/itex]