Answer
$0$
Work Step by Step
Green's Theorem states that:
$\oint_CP\,dx+Q\,dy=\iint_{D}(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})dA$
The parametric representation for the boundary of the ellipse can be written as: $x= \sqrt 2 \cos \theta; y= \sin \theta$
and {$(r, \theta) \in D| 0 \leq r \leq 1; 0 \leq \theta \leq 2 \pi$}
We can work out the integrand of the double integral as follows:
$\oint_C y^4 dx+2xy^3 dy=\iint_{D}(\dfrac{\partial (2xy^3 )}{\partial x}-\dfrac{\partial (y^4 ) }{\partial y})dA=\iint_{D} -2y^3 dA$
Rewriting the double integral as an iterated integral and solving, we get:
$\iint_{D} -2y^3 dA=\int_{0}^{2 \pi} \int_{0}^{1} -2 (r \sin \theta)^3 (r \sqrt 2) dr d \theta$
Consider $I=\int_{0}^{2 \pi} \int_{0}^{1} -2 (r \sin \theta)^3 (r \sqrt 2) dr d \theta \\ =-2 \sqrt 2 (\int_{0}^{2 \pi} \sin^3 \theta d\theta ) ( \int_0^1 r^4 dr)$
Since, $\int_{0}^{2 \pi} \sin^3 \theta d\theta =0 $
Therefore, $\oint_C y^4 dx+2xy^3 dy=-2 \sqrt 2 \times (0) \times \int_0^1 r^4 dr=0$