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Math Help - Calculus III problem

  1. #1
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    Exclamation Calculus III problem

    Hi, so tomorrow's pretty much the last day of the summer semester for me and I have a homework assignment due tomorrow that's confusing me sooo much..

    Im here asking for some help in direction on how to do these problems so that I can find the solution by myself... I would really really appreciate any help anyone could provide ...

    I am not sure how to represent the domain as a function, and if anyone could point me in the right direction, I'm sure I could get the answer.

    The Problem -



    I would really really really appreciate any help at all. I need to know how to represent the domain as a function.
    Last edited by mr fantastic; July 28th 2010 at 10:03 PM.
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  2. #2
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    Don't think of the domain as one big function. Think of it as 3 parts as in the descrption:

    C1: line from (0,0) to (1,0)
    C2: circular arc from (1,0) to the line x=y. What is x in this case?
    C3: line from (x,x) to (0,0)

    If you don't know how to parametrize lines, look in your book somewhere around parametric equations of a line.
    The circular arc you can just parametrize using sines and cosines.

    I hope this helps!
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  3. #3
    Junior Member bondesan's Avatar
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    I think we should go for Green's Theorem.

    \displaystyle{\int_{\partial R} P dx + Q dy = \int\int_{R}\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dxdy}

    So, lets do it:

    Let have P = \sin x - 6x^2y and Q=3xy^2-x^3.

    \frac{\partial Q}{\partial x} = 3y^2 - 3x^2

    \frac{\partial P}{\partial y} = - 6x^2

    \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = 3y^2 + 3x^2 (How convenient!!)

    Now we have to determine the region bounded by the curve C. I've made a sketch:

    Calculus III problem-green2.jpg

    You see that we're "walking" counter-clockwise - thus we should have \int\int_{R} 3y^2 + 3x^2~dxdy. There would be a negative signal if we walked clockwise.

    Well. We should determine the limits of integration. I mentioned above the convenience because this integral is asking so much that we change it for polar coordinates - therefore we can easily find the limits of integration (I think the sketch speaks by itself). So we put:

    x = r cos t and y=r sin t.

    Our angle will vary from 0 to \pi/2 + \pi/4 = 3\pi/4 and the radius will vary from 0 to 1 (unit circle). This will lead us to:

    Ah, before, we can't forget the Jacobian (I will not show this, but you should know how its done), so  dx dy = r dr dt. Finally, the integral becomes:

    \displaystyle{\int\int_{R} 3 (y^2 + x^2)~dxdy = \int_{0}^{3\pi/4}\int_{0}^{1} 3 r^2\cdot r~drdt because x^2+y^2 = r^2 by our change.
    Last edited by bondesan; July 28th 2010 at 08:45 PM. Reason: Wrong figure
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  4. #4
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    It says counter-clockwise in the question!
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  5. #5
    Junior Member bondesan's Avatar
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    Oh, my mistake, sorry. I'll correct it.
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  6. #6
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    Thank you so much for the quick replies. I appreciate it a looooot! I will try it out and see what I can come up with. And keep you guys updated.
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  7. #7
    Junior Member bondesan's Avatar
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    Already done, I guess you can follow now.
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