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Thread: 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: $\displaystyle -\sqrt{9-x^2} < y < \sqrt{9-x^2}$ is equivalent to the interior of the circle $\displaystyle x^2 + y^2 = 9$.
    $\displaystyle 0 < x < 3$ 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 $\displaystyle dx \, dy \rightarrow r \, dr\, d\theta$, 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; Oct 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 $\displaystyle 0 < \theta < \pi$ 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 $\displaystyle x = r \cos \theta$ and $\displaystyle y = r \sin \theta$ into $\displaystyle 3y - x$ and get $\displaystyle r \cos \theta + 3r \sin \theta$ ....?

    By the way, don't bump.
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