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    My notes say this:

    "Sometimes the order of the integration will determine whether or not it is necessary to split the region of integration..."

    But goes into no more detail?

    In which situations would you not need to split the region in one order, but you would in the other?

    Thankyou.
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    Consider the region D defined by 1-x^2 <= y <= 2(1-x^2), -1<=x<=1. (This is basically a region enclosed by two parabolas and the x-axis).

    If you do \displaystyle \int_{x=-1}^1 \int_{y=1-x^2}^{2(1-x^2)} f(x,y) \,dy\,dx you can do it in one go.

    But if you integrate the other way around, there are two 'ranges' of x value for each value of y. (Specifically, \sqrt{1-y}\le x \le \sqrt{1-y/2} and -\sqrt{1-y/2}\le x \le -\sqrt{1-y}), so you'd need to split into two separate regions.
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    (Original post by DFranklin)
    Consider the region D defined by 1-x^2 <= y <= 2(1-x^2), -1<=x<=1. (This is basically a region enclosed by two parabolas and the x-axis).

    If you do \displaystyle \int_{x=-1}^1 \int_{y=1-x^2}^{2(1-x^2)} f(x,y) \,dy\,dx you can do it in one go.

    But if you integrate the other way around, there are two 'ranges' of x value for each value of y. (Specifically, \sqrt{1-y}\le x \le \sqrt{1-y/2} and -\sqrt{1-y/2}\le x \le -\sqrt{1-y}), so you'd need to split into two separate regions.
    I see.

    So any situation where, if you change the order, and the limits of integration can subsequently have 1 value or another, then you must split into two regions and sum them?
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    Pretty much. If you draw a sketch, it should be pretty obvious what's going on - it's just a hassle to do sketches on TSR.
 
 
 
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