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Math Help - changing the order of integration

  1. #1
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    changing the order of integration

    So here I got such integral. How to change the order of integration? I am not asking you for the answer, but what should be the first step?

    \int_{-1}^{1}dy  \int_{{y^2 \over 2}}^{ \sqrt{3-y^2}} f(x,y)dx
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  2. #2
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    Quote Originally Posted by jacek View Post
    So here I got such integral. How to change the order of integration? I am not asking you for the answer, but what should be the first step?

    \int_{-1}^{1}dy  \int_{{y^2 \over 2}}^{ \sqrt{3-y^2}} f(x,y)dx
    I assume this is actually

    \int_{-1}^{1}{\int_{\frac{y^2}{2}}^{\sqrt{3 - y^2}}{f(x, y)\,dx}\,dy}.


    It's important to realise that the terminals actually give you a boundary for x and y.


    So \frac{y^2}{2}\leq x \leq \sqrt{3 - y^2}

    and -1 \leq y \leq 1.


    You should be able to rearrange the inequalities so that you can change the order of integration.
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  3. #3
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    so the 'lines' of this are will be:
    y=-1
    y=1
    x=(y^2)/2
    x= \sqrt{(3-y^2)}?
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  4. #4
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    It really helps to draw a picture for this one. If you change the order of integration, you will need two integrals instead of one.
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  5. #5
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    Yep, but should I draw it according to these 'lines' above?
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  6. #6
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    Quote Originally Posted by jacek View Post
    Yep, but should I draw it according to these 'lines' above?
    Yes. A slight correction: if you use only integrals to describe the area and you switch the order of integration, you will need three integrals, not two, although one of these is a rectangle.
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  7. #7
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    That's ok, my teacher did it once only with integrals.

    Could you tell me what's the easiest way to draw x=\sqrt{(3-y^2)} without calculator?
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  8. #8
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    Quote Originally Posted by jacek View Post
    That's ok, my teacher did it once only with integrals.

    Could you tell me what's the easiest way to draw x=\sqrt{(3-y^2)} without calculator?
    Square both sides and you get x^2 = 3 - y^2, or x^2 + y^2 = 3, which is a circle centered at the origin with radius \sqrt{3}. Note that x=\sqrt{(3-y^2)} is only part of this circle. The best way to draw it would be with pencil and compass.
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  10. #10
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    Quote Originally Posted by jacek View Post
    Something like that. Is that the best parabola you can draw?
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  11. #11
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    Hehe, no! I can draw much better, but this is just auxiliary drawing, isn't it?

    So, I need to divide it into three parts - part of circle on the right, rectangle in the middle, and this parabola part on the the left?
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  12. #12
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    Quote Originally Posted by jacek View Post
    Hehe, no! I can draw much better, but this is just auxiliary drawing, isn't it?

    So, I need to divide it into three parts - part of circle on the right, rectangle in the middle, and this parabola part on the the left?
    Yes, that is correct.
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  13. #13
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    Aha, thanks!

    What about such integral?

    \int_{-1}^{1}dx  \int_{0}^{ \sqrt{1-y^2}} f(y)

    How can I change it? Because it's only a rectangle...
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  14. #14
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    Anyone?
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