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can someone please help me understand how to find the 'new' limits for double integration. I know that you have to split up the area and fix x or y. If you can go through an example with me then I would be grateful as I keep getting this wrong. We have only been taught how to work out the area of a triangle and so a quadrilateral makes no sense to me.... Btw please don’t give 'genera;' advice because I seriously won’t get it. I have looked in my books/lecture notes but it’s all just general theory which isn’t helpful.

Example question: By making an appropriate substitution, find the area of D and check the substitution is a 1-1 transformation by finding the inverse explicitly.

$$ \iint_D \exp{[xy(x-y)]} (x^2-y^2) {\rm d}x\, {\rm d}y$$

Here is the diagram: enter image description here Thanks for the help!

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You've asked quite a number of questions but haven't accepted any of the answers. Were none of them satisfactory? Also, the image requires two clicks to get to -- you have enough reputation to include it directly in the question. – joriki May 25 '11 at 11:22
There seem to be two different questions here -- one is to calculate the area of $D$, which would involve integrating $1$ over $D$, and the other is to integrate $\exp{[xy(x-y)]} (x^2-y^2)$ over $D$ -- which one are you trying to solve? Or both? – joriki May 25 '11 at 11:29
Sorry - I'm new here! How do you accept an answer? How do you attach a pic directly? I never knew I could. And I just realised that my post did not paste properly... I will adjust this tomorrow! So sorry about that. – user4645 May 26 '11 at 0:33
@user4645: You can accept an answer by clicking on the checkmark symbol next to it. You can attach an image by clicking on the corresponding icon (the one next to the "binary" symbol) in the toolbar above the edit textfield. – joriki May 26 '11 at 4:20
checkmark symbol???? ive made a new post instead to clarify my problems. i think its easier that way. someone please close this thread... – user4645 May 27 '11 at 15:48

You don't have to split up the area. The boundary curves of $D$ are given by constant values of $x-y$ and $xy$. Thus, it makes sense to transform to new variables $u=x-y$ and $v=xy$; then the integration region will turn into an axis-parallel rectangle, and you can just perform the integration without further worrying about the integration limits.

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