Help evaluating a double integral? (maths)

[titaniumx3]

I'm really stumped in this question, in particular, how to deal with the region of integration. The question itself is pretty simple, here it goes:

Evaluate:

int(int(exp(x+y))).dA

with the region of integration: |x|+|y|<= 1

(note: |x|,|y| denote the absolute values)

Use the substitution x=(u+v)/2 and y=(u-v)/2

Using the given substitution the jacobian turns out to be -1/2 hence the integral becomes:

int(int(exp(u))).1/2.du.dv

But what do you do with the region of integration; how do you figure out the limits?

Any help would be greatly appreciated!

 

[Vonhelmet]

|X| + |Y| <= 1

Um...

Basically, it'll be a diamond, 2 units from point to point, centered on the origin. Not sure how to prove that mathematically.

It'll look like something like this:

      1
     /|\
    / | \
   /  |  \
-1/__0|___\1
  \   0   /
   \  |  /
    \ | /
     \|/
      -1

 

[titaniumx3]

Thanks for the reply! Right I get that, but so I would I simply split the integral up? How would you go about writing that out in terms of the substitutions in u,v?

 

[Vonhelmet]

Thanks for the reply! Right I get that, but so I would I simply split the integral up? How would you go about writing that out in terms of the substitutions in u,v?

Not sure about that side of things... thought you were just asking about how the limits looked...

Sorry! Don't have my degree notes here with me, so can't even look there.

 

[Tokenbrit]

If you want to change the limits from the variables x,y to u,v you already know these limits implicitly at least:
|(u+v)/2| + |(u-v)/2|<=1

Now if you want you can solve for u,v to get the explicit limits.

 

[titaniumx3]

This is still confusing the hell out of me, how on earth do you find the explicit limits from that?

 

[Arcade Fire]

You need to look at the equation |x|+|y| <= 1 in the four quadrants:

1. x>0, y>0,
2. x<0, y>0,
3. x<0, y<0,
4. x>0, y<0.

For example, in quadrant 1 you have |x|=x and |y|=y, so the limit is

x+y <= 1
(u+v)/2 + (u-v)/2 <= 1
u <= 1.

In quadrant 2 you have |x|=-x and |y|=y, so the limit is

-x+y <= 1
-(u+v)/2 + (u-v)/2 <= 1
-v <= 1
v >= -1.

And similarly the the other two quadrants. You should find that you end up integrating between -1 <= u <= 1 and -1 <= v <= 1.

 

[Tokenbrit]

Arr they be the limits. Seems very redundant to me to perform the change of variables though, would have been easier just to perform the double integral wrt x,y than mess around calculating Jacobians and limits .

 

[DaveF]

You need to look at the equation |x|+|y| <= 1 in the four quadrants: ~snip~

Alternatively, express u,v back in terms of x and y: u = x+y, v=x-y. So a line of constant u will be a line of constant x+y, which 'just happens' to be two of the lines in the region you've already sketched. Substitute in a couple of the limit points (e.g. (1,0) and (-1,0)) and you see -1<=u<=1. Similarly for v.

That's probably not too helpful when put in words; it works much better with a diagrom. You'd just label the 4 sides of the 'diamond' as x+y = 1, x+y = -1, x-y = 1, x-y = -1 and then you can just write down the limits.

This might not work too well for a real-world problem, but the questions you get asked at university tend to be artificial, so if you have an awkward shape for your original integral, you get a nice one for the final one. Once you have a bit of practice, often you can just 'see' how it's going to work out.