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Math Help - Convert double integral to polar

  1. #1
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    Convert double integral to polar

    Hi Guys / girls

    I am going through my home work for uni and i have to calculate the integral of the following by transforming to polar coordinates. This fine except i just cant get my head around how to do this.

    The problem is -//(3y-x)dydx 0<x<3, -sqrt(9-x^2) < y < sqrt(9-x^2)

    Thanks in advance
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  2. #2
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    Quote Originally Posted by bamptom View Post
    Hi Guys / girls

    I am going through my home work for uni and i have to calculate the integral of the following by transforming to polar coordinates. This fine except i just cant get my head around how to do this.

    The problem is -//(3y-x)dydx 0<x<3, -sqrt(9-x^2) < y < sqrt(9-x^2)

    Thanks in advance
    Have you drawn the region of integration?

    Note: -\sqrt{9-x^2} < y < \sqrt{9-x^2} is equivalent to the interior of the circle x^2 + y^2 = 9.
    0 < x < 3 means you want the right half of this circle ......

    Can you define this region using polar coordinates? That will give you the integral terminals.

    And you know dx \, dy \rightarrow r \, dr\, d\theta, right?
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  3. #3
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    Are they 0<Theta<pi and 0<r<3 ??

    And does (3y-x)dydx >>>>> // (rcostheta + 3rsintheta)rdrdtheta

    I really do appreciate your help!
    Last edited by bamptom; October 23rd 2008 at 02:04 AM.
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  4. #4
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    ??

    Can anyone tell me if im on the right track here?
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  5. #5
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    Quote Originally Posted by bamptom View Post
    Are they 0<Theta<pi and 0<r<3 ??

    [snip]
    No. Have you drawn the region like I suggested? What's the polar coordinates of (0, -3) and (0, 3) .....? Will 0 < \theta < \pi give you those points .....?

    Quote Originally Posted by bamptom View Post
    [snip]
    And does (3y-x)dydx >>>>> // (rcostheta + 3rsintheta)rdrdtheta

    I really do appreciate your help!
    No. How can you substitute x = r \cos \theta and y = r \sin \theta into 3y - x and get  r \cos \theta + 3r \sin \theta ....?

    By the way, don't bump.
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