Answer
Green's Theorem has been verified.
Work Step by Step
Green's Theorem states that:
$\oint_C M \,dx+ N \,dy=\iint_{D}(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y})dA$
We will use polar coordinates as follows: $x=r \cos \theta; y= r \sin \theta$
We need to work out with the line integral and evaluate the integrand of the double integral as follows:
$\oint_CP\,dx+Q\,dy=\int_{C} (2x-x^3y^5) \ dx +(x^3 y^8) \ dy$
Set $ \cos \theta = x$ and $2 \sin \theta =y$
$\int_{C} (2x-x^3y^5) \ dx +(x^3 y^8) \ dy =\int_{0}^{2 \pi} 2 \cos \theta \times \sin \theta (16 \cos^2 \theta \sin^4 \theta+256 \cos^3 \theta \sin^7 \theta-1) d \theta$
Use calculator, so we have: $\int_{0}^{2 \pi} 2 \cos \theta \times \sin \theta (16 \cos^2 \theta \sin^4 \theta+256 \cos^3 \theta \sin^7 \theta-1) d \theta = 7 \pi ...(a)$
Now, the double integral is: $\iint_{D}(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})dA=\iint_{D}(\dfrac{\partial (x^3 y^8)}{\partial x}-\dfrac{\partial (2x-x^3y^5)}{\partial y}) \ dA =\int_{-1}^1 \int_{2 \sqrt {1-x^2}}^{\sqrt{1-x^2}} 3x^2 y^8+5x^3y^4 dy dx$
Use calculator , so we have:$\int_{-1}^1 \int_{2 \sqrt {1-x^2}}^{\sqrt{1-x^2}} 3x^2 y^8+5x^3y^4 dy dx=7 \pi ....(b)$
So, from equations (a) and (b) it has been seen that the Green's Theorem has been verified.
