# Use Green's Theorem to evaluate oint_C(x^2+y)dx-(3x+y^3)dy Where c is the ellipse x^2+4y^2=4

Use Greens
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Step 1
It is required to calculate the value of the integral using green’s theorem:

Step 2
${\oint }_{C}Pdx+Qdy=\int {\int }_{R}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA$
Step 3
So,
${\oint }_{C}\left({x}^{2}+y\right)dx-\left(3x+{y}^{3}\right)dy=\int {\int }_{R}\frac{\partial }{\partial x}\left(-\left(3x+{y}^{3}\right)\right)-\frac{\partial }{\partial y}\left({x}^{2}+y\right)$
$=\int {\int }_{R}\left(-3-1\right)dA$
$=\int {\int }_{R}\left(-4\right)dA$
Step 4
${\oint }_{C}\left({x}^{2}+y\right)dx-\left(3x+{y}^{3}\right)dy=\int {\int }_{R}\left(-4\right)dA={\int }_{-1}^{1}{\int }_{-2}^{2}\left(-4\right)dxdy$
$=\left(-4\right){\int }_{-1}^{1}{\left[x\right]}_{-2}^{2}dy$
$=\left(-4\right){\int }_{-1}^{1}\left(2-\left(-2\right)\right)dy$
$=\left(-4\right){\int }_{-1}^{1}4dy$
$=\left(-16\right){\left(y\right)}_{-1}^{1}$
$=-16\left(1-\left(-1\right)\right)$
$=-16\left(2\right)$
${\oint }_{C}\left({x}^{2}+y\right)dx-\left(3x+{y}^{3}\right)dy=-32$

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