Answer
$0$
Work Step by Step
Green's Theorem states that:
$\oint_C M \,dx+ N \,dy=\iint_{D}(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y})dA$
The parametric representation for the boundary of ellipse can be expressed as: $x= \sqrt 2 \cos \theta; y= \sin \theta$
and {$(r, \theta) \in D| 0 \leq r \leq 1; 0 \leq \theta \leq 2 \pi$}
The integrand of the double integral becomes:
$\oint_C y^4 dx+2xy^3 dy=\iint_{D}(\dfrac{\partial (2xy^3 )}{\partial x}-\dfrac{\partial (y^4 ) }{\partial y})dA \\ =\iint_{D} -2y^3 dA\\=\int_{0}^{2 \pi} \int_{0}^{1} -2 (r \sin \theta)^3 (r \sqrt 2) dr d \theta$
Consider $I=\int_{0}^{2 \pi} \int_{0}^{1} -2 (r \sin \theta)^3 (r \sqrt 2) dr d \theta \\ =-2 \sqrt 2 (\int_{0}^{2 \pi} \sin^3 \theta d\theta ) \times ( \int_0^1 r^4 dr)$
Since, $\int_{0}^{2 \pi} \sin^3 \theta d\theta =0 $
So, $\oint_C y^4 dx+2xy^3 dy=-2 \sqrt 2 \times (0) \times \int_0^1 r^4 dr=0$
