# [SOLVED] Applying Green's theorem

• Jun 15th 2009, 04:04 PM
arbolis
[SOLVED] Applying Green's theorem
I must apply Green's theorem in order to evaluate the following line integral : $\displaystyle I=\int_C x^2ydx-3y^2dy$ where $\displaystyle C$ is the circumference $\displaystyle x^2+y^2=1$.
My attempt : $\displaystyle \int_C x^2ydx-3y^2dy=\iint_{x^2+y^2=1} -x^2dxdy$.
$\displaystyle r(t)=(\cos t, \sin t) \Rightarrow x=\cos t$, $\displaystyle y=\sin t$ so $\displaystyle I=\int _0^1 \int _{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} -\cos^2 (t) (-\sin (t))dt \cdot \cos (t) dt$.
Oh oh... $\displaystyle dt\cdot dt = dt^2$? lol.
Where did I made an error?
• Jun 15th 2009, 04:14 PM
NonCommAlg
Quote:

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

the double integral is over the region $\displaystyle x^2+y^2 \leq 1$ not over the curve $\displaystyle x^2+y^2=1.$ change to polar coordinates:

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