# Use Green's Theorem to find int_C vec(F) * d vec(r) where vec(F) = <<y^3, - x^3>> and C is the circle x^2+y^2=3.

Use Green's Theorem to find ${\int }_{C}\stackrel{\to }{F}\cdot d\stackrel{\to }{r}$ where $\stackrel{\to }{F}=⟨{y}^{3},-{x}^{3}⟩$ and C is the circle ${x}^{2}+{y}^{2}=3$.
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Step 1
The given vector is
$F={y}^{3}i-{x}^{3}j$
Step 2
$I={\int }_{C}\stackrel{\to }{F}\cdot d\stackrel{\to }{r}$
but d $\stackrel{\to }{r}=dx\stackrel{^}{i}+dy\stackrel{^}{j}$
$I={\int }_{C}\left({y}^{3}\stackrel{^}{i}-{x}^{3}\stackrel{^}{j}\right).\left(dx\stackrel{^}{i}+dy\stackrel{^}{j}\right)$
$={\int }_{C}\left({y}^{3}dx-{x}^{3}dy\right)$
The given circle
${x}^{2}+{y}^{2}=3$
${x}^{2}+{y}^{2}={\left(\sqrt{3}\right)}^{2}$
$radius=\sqrt{3}$
Apply Green Theorem
${\int }_{C}\left(Mdx+Ndy\right)=\int \int \left(\frac{\partial N}{\partial x}-\frac{DelM}{\partial y}\right)dxdy$
$⇒I=\int \int \left(-3{x}^{2}-3{y}^{2}\right)dxdy$
$=\int \int -3\left({x}^{2}+{y}^{2}\right)dxdy$
$=-3\int \int \left({x}^{2}+{y}^{2}\right)dxdy$
$=-9\int \int dxdy\because {x}^{2}+{y}^{2}=3$
$=-9x$ Area of circle
$=-9x\pi {r}^{2}$
$=-9x\pi x{\left(\sqrt{3}\right)}^{2}$
$I=-27\pi$