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Math Help - Polar equivalent integral

  1. #1
    s7b
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    Polar equivalent integral

    Hi,

    I'm supposed to change this cartesian integral into a polar integral.

    Cartesian form:
    (integral from y=0 to y=2)(integral from x=-(sqrt(1-(y-1)^2)) to x=0) of xy^2 dxdy

    I got the polar equivalent to be:

    (integral from theta=-pi/2 to pi/2)(integral from r=0 to r=1) of r^4(costheta)(sin^2theta) drdtheta

    Have I done this correctly??
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  2. #2
    Flow Master
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    Quote Originally Posted by s7b View Post
    Hi,

    I'm supposed to change this cartesian integral into a polar integral.

    Cartesian form:
    (integral from y=0 to y=2)(integral from x=-(sqrt(1-(y-1)^2)) to x=0) of xy^2 dxdy

    I got the polar equivalent to be:

    (integral from theta=-pi/2 to pi/2)(integral from r=0 to r=1) of r^4(costheta)(sin^2theta) drdtheta

    Have I done this correctly??
    No. Your new integral terminals are wrong.

    Note that the area is the left half of the circle (y - 1)^2 + x^2 = 1 so you need to express that circle in polar form.

    Note that the left half of the circle is gven by \frac{\pi}{2} \leq \theta \leq \pi.
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  3. #3
    s7b
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    When I take the integral after changing the limits you said I end up with negative 1/15. Shouldn't the answer be positive??
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  4. #4
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    Quote Originally Posted by s7b View Post
    When I take the integral after changing the limits you said I end up with negative 1/15. Shouldn't the answer be positive??
    Excuse me ....? Where did I say that??

    Re: Expressing the circle in polar form. Reading this thread might help: http://www.mathhelpforum.com/math-he...variables.html
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  5. #5
    MHF Contributor

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    S7b, the integral must be negative. It is the integral of xy^2 and is taken over a region where x is always negative: x ranges from -\sqrt{1- (y-1)^2} up to 0 for every y.
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