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  1. #1
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    Question Find the area bounded by...

    Find the area bounded by,

    (x^{2}+y^{2})^{3} = 4a^{2}x^{2}y^{2}

    I'm a little confused by the question itself. What are they talking about? Is it a surface? It it simply something in the x-y plane? How do I get this one started?

    Thanks again!
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by jegues View Post
    Find the area bounded by,

    (x^{2}+y^{2})^{3} = 4a^{2}x^{2}y^{2}

    I'm a little confused by the question itself. What are they talking about? Is it a surface? It it simply something in the x-y plane? How do I get this one started?

    Thanks again!
    Use the polar coordinates

    \displaystyle{(x^2+y^2)^3=4a^2x^2y^2}

    \displaystyle{x=\rho\cos\varphi,~y=\rho\sin\varphi  }

    \displaystyle{\rho^6=4a^2\rho^4\cos^2\varphi\sin^2  \varphi}

    \displaystyle{\rho^2=4a^2\cos^2\varphi\sin^2\varph  i}

    \displaystyle{\rho^2=a^2\sin^22\varphi}

    \displaystyle{A=\frac{1}{2}\int\limits_\alpha^\bet  a\rho^2\,d\varphi=\frac{a^2}{2}\int\limits_0^{2\pi  }\sin^22\varphi\,d\varphi=\ldots=\frac{a^2\pi}{2}}


    P.S. \rho=a\sin2\varphi is the equation of the polar rose with four petals (see wikipedia).
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  3. #3
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    Quote Originally Posted by DeMath View Post
    Use the polar coordinates

    \displaystyle{(x^2+y^2)^3=4a^2x^2y^2}

    \displaystyle{x=\rho\cos\varphi,~y=\rho\sin\varphi  }

    \displaystyle{\rho^6=4a^2\rho^4\cos^2\varphi\sin^2  \varphi}

    \displaystyle{\rho^2=4a^2\cos^2\varphi\sin^2\varph  i}

    \displaystyle{\rho^2=a^2\sin^22\varphi}

    \displaystyle{A=\frac{1}{2}\int\limits_\alpha^\bet  a\rho^2\,d\varphi=\frac{a^2}{2}\int\limits_0^{2\pi  }\sin^22\varphi\,d\varphi=\ldots=\frac{a^2\pi}{2}}


    P.S. \rho=a\sin2\varphi is the equation of the polar rose with four petals (see wikipedia).
    I have one quick question regarding this line,

    \displaystyle{A=\frac{1}{2}\int\limits_\alpha^\bet  a\rho^2\,d\varphi<br />

    Where does this come from? Is this a known formula for area or something?
    Last edited by jegues; November 27th 2010 at 07:38 AM.
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by jegues View Post
    I have one quick question regarding this line,

    \displaystyle{A=\frac{1}{2}\int\limits_\alpha^\bet  a\rho^2\,d\varphi<br />

    Where does this come from? Is this a known formula for area or something?
    Yes

    See Integral calculus (Area) Polar coordinate system - Wikipedia, the free encyclopedia
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