Answer
$$0$$
Work Step by Step
Write the parametric representation for the boundary of the ellipse as: $x= \sqrt 2 \cos \theta; y= \sin \theta$
and {$(r, \theta) \in D| 0 \leq r \leq 1; 0 \leq \theta \leq 2 \pi$}
Green's Theorem states that:
$\oint_C A\,dx+B \,dy=\iint_{D}(\dfrac{\partial B}{\partial x}-\dfrac{\partial A}{\partial y}) dA $
We need to work out the integrand of the double integral as follows:
$\oint_C y^4 dx+2xy^3 dy=\iint_{D}(\dfrac{\partial (2xy^3 )}{\partial x}-\dfrac{\partial (y^4 ) }{\partial y})dA=\iint_{D} -2y^3 dA$
Now, we will rewrite 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 $I=\int_{0}^{2 \pi} \int_{0}^{1} -2 (r \sin \theta)^3 (r \sqrt 2) dr d \theta =-2 \sqrt 2 \times (\int_{0}^{2 \pi} \sin^3 \theta d\theta ) ( \int_0^1 r^4 dr)$
Since, $\int_{0}^{2 \pi} \sin^3 \theta d\theta =0 $
Thus, we get: $\oint_C y^4 dx+2xy^3 dy=-(2 \sqrt 2) (0) (\int_0^1 r^4 dr)=0$
