Answer
$\int_Cy^{3}dx+x^{2}dy=\frac{4}{15}$
Work Step by Step
We can parameterize the parabola $x=1-y^{2}$ as
$C:x=1-y^{2},y=t$
As we move from $(0,-1)$ to $(0,1)$ , $t$ increases from $-1$ to $1$.
Therefore, we will integrate from -1 to 1.
Note that $dx=-2tdt$ and $dy=dt$
Therefore,
$\int_Cy^{3}dx+x^{2}dy=\int_{-1}^{1}(t)^{3}(-2tdt)(1-t^{2})^{2}dt$
$=\int_{-1}^{1}(-2t^{4}+t^{4}-2t^{2}+1)dt$
$=\int_{-1}^{1}(-t^{4}-2t^{2}+1)dt$
$=[(\frac{-t^{5}}{5}-\frac{2t^{3}}{3}+t)]_{-1}^{1}$
$=2(\frac{-3}{15}-\frac{10}{15}+\frac{15}{15})$
Hence,
$\int_Cy^{3}dx+x^{2}dy=\frac{4}{15}$
