[SOLVED] Applying Green's theorem

**arbolis** · June 15th 2009, 04:04 PM

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?

**NonCommAlg** · June 15th 2009, 04:14 PM

Originally Posted by arbolis

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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