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Math Help - [SOLVED] Applying Green's theorem

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Applying Green's theorem

    I must apply Green's theorem in order to evaluate the following line integral : I=\int_C x^2ydx-3y^2dy where C is the circumference x^2+y^2=1.
    My attempt : \int_C x^2ydx-3y^2dy=\iint_{x^2+y^2=1} -x^2dxdy.
    r(t)=(\cos t, \sin t) \Rightarrow x=\cos t, y=\sin t so I=\int _0^1 \int _{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} -\cos^2 (t) (-\sin (t))dt \cdot \cos (t) dt.
    Oh oh... dt\cdot dt = dt^2? lol.
    Where did I made an error?
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    Quote Originally Posted by arbolis View Post
    I must apply Green's theorem in order to evaluate the following line integral : I=\int_C x^2ydx-3y^2dy where C is the circumference x^2+y^2=1.
    My attempt : \int_C x^2ydx-3y^2dy=\iint_{x^2+y^2=1} -x^2dxdy.
    r(t)=(\cos t, \sin t) \Rightarrow x=\cos t, y=\sin t so I=\int _0^1 \int _{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} -\cos^2 (t) (-\sin (t))dt \cdot \cos (t) dt.
    Oh oh... dt\cdot dt = dt^2? lol.
    Where did I made an error?
    the double integral is over the region x^2+y^2 \leq 1 not over the curve x^2+y^2=1. change to polar coordinates:

    -\int \int_{x^2+y^2 \leq 1} x^2 \ dx dy = -\int_0^{2 \pi} \int_0^1 r^3 \cos^2 \theta \ dr d \theta= -\frac{1}{4} \int_0^{2 \pi} \cos^2 \theta \ d\theta=\frac{-\pi}{4}.
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