Answer
Green's Theorem has been verified.
Work Step by Step
We can parameterize the curve as $x= \cos \theta; y= 2 \sin \theta$ and $ 0 \leq \theta \lt 2 \pi $
Green's Theorem states that:
$\oint_CP\,dx+Q\,dy=\iint_{D}(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})dA$
The line integral is:
$\oint_CP\,dx+Q\,dy=\int_{C} (2x-x^3y^5) \ dx +(x^3 y^8) \ dy$
Substitute $x= \cos \theta; y= 2 \sin \theta$
Then, we have:
$\int_{C} (2x-x^3y^5) \ dx +(x^3 y^8) \ dy =\int_{0}^{2 \pi} 2 \cos \theta \sin \theta (16 \cos^2 \theta \sin^4 \theta+256 \cos^3 \theta \sin^7 \theta-1) d \theta$
By using a calculator we have:
$\int_{0}^{2 \pi} 2 \cos \theta \sin \theta (16 \cos^2 \theta \sin^4 \theta+256 \cos^3 \theta \sin^7 \theta-1) d \theta = 7 \pi ...(1)$
Now, the double integral is:
$\iint_{D}(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})dA=\iint_{D}(\dfrac{\partial (x^3 y^8)}{\partial x}-\dfrac{\partial (2x-x^3y^5)}{\partial y})dA$
or, $=\int_{-1}^1 \int_{2 \sqrt {1-x^2}}^{\sqrt{1-x^2}} 3x^2 y^8+5x^3y^4 dy dx$
By using a calculator, we have:
$\int_{-1}^1 \int_{2 \sqrt {1-x^2}}^{\sqrt{1-x^2}} 3x^2 y^8+5x^3y^4 dy dx=7 \pi ....(2)$
Hence, from equations (1) and (2), Green's Theorem has been verified.
