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    I'm having difficulties actually finding out the exact domain of integration for both parts b) and c)

    For b, I got x -> from -1 to 1 and y-> from 0 to x^2+y^2 but I'm unsure if that's correct.

    Help would be much appreciated
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    (Original post by TwiMaster)
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    I'm having difficulties actually finding out the exact domain of integration for both parts b) and c)

    For b, I got x -> from -1 to 1 and y-> from 0 to x^2+y^2 but I'm unsure if that's correct.

    Help would be much appreciated
    I disagree for b). Have you sketched the region defined by those inequalities? If I'm correct, it all lies in the first quadrant, so you wont need -ve x at all.

    Your limits for x sweep the whole unit circle - your limits for y are ill-defined, since when you integrate over y, you must remove any dependence on y, in order to end up with a definite integral. You will be writing:

    \displaystyle \int_{x=-1}^{x=1} \big(\int_{y=0}^{y=x^2+y^2} f(x,y) \ dy \big) \ dx

    which will leave a y variable to be integrated against dx.
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    (Original post by TwiMaster)
    I'm having difficulties actually finding out the exact domain of integration for both parts b) and c)

    For b, I got x -> from -1 to 1 and y-> from 0 to x^2+y^2 but I'm unsure if that's correct.

    Help would be much appreciated
    b) Well the limits on y are not going to be a function of y, so that's not correct.
    Try sketching the desired area to begin with.


    c) Notice that your first constraint is a perfect square on one side. In each case you can express your contraint in terms of x^2+y^2, from which it should be easy to sketch the domain.

    Edit: ninja'd
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    (Original post by ghostwalker)
    b) Well the limits on y are not going to be a function of y, so that's not correct.
    Try sketching the desired area to begin with.


    c) Notice that your first constraint is a perfect square on one side.
    Perfect square?

    Edit: ninja'd
    Well, I have to lie in wait these days, and pounce immediately a half-decent question comes up, there's so few of them.
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    (Original post by atsruser)
    Perfect square?
    x^4+y^4+2x^2y^2 = (x^2+y^2)^2

    Or am I being dense?
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    (Original post by ghostwalker)
    x^4+y^4+2x^2y^2 = (x^2+y^2)^2

    Or am I being dense?
    No, it's me - I didn't notice that you were talking about part c).
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    (Original post by atsruser)
    I disagree for b). Have you sketched the region defined by those inequalities? If I'm correct, it all lies in the first quadrant, so you wont need -ve x at all.

    Your limits for x sweep the whole unit circle - your limits for y are ill-defined, since when you integrate over y, you must remove any dependence on y, in order to end up with a definite integral. You will be writing:

    \displaystyle \int_{x=-1}^{x=1} \big(\int_{y=0}^{y=x^2+y^2} f(x,y) \ dy \big) \ dx

    which will leave a y variable to be integrated against dx.
    For b.) I don't quite understand how it all lies in the first quadrant, perhaps I've incorrectly sketched the region because of the inequalties
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    (Original post by TwiMaster)
    For b.) I don't quite understand how it all lies in the first quadrant, perhaps I've incorrectly sketched the region because of the inequalties
    Can you put up your sketch?

    x^2+y^2 \le 1 defines the boundary and interior of the unit circle.

    y \ge 1-x is the half-plane with boundary above y=1-x, which line intersects the y-axis at (0,1) and the x-axis at (1,0).

    Does the intersection of those two not lie in the first quadrant?
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    (Original post by TeeEm)
    similar questions to yours to help you.
    ... it should be obvious from the solution what the question is

    Attachment 609076

    Ah, this is great! Thanks for this
    It's pretty difficult to find worked solutions at uni level like this
 
 
 
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