Let $C$ be the border of the surface $S$ with equation $x^{2} + y^{4} + z^{4} = 1$ and $v= (3zx^{2}, 12\sin y, x\cos z^{3} )$. Then calculate $\int_{C}^{}v dt$.
I know that this is a line integral. I tried to find its border. For $z=0:$ I got $\partial S: x^{2} + y^{4}=1$ I took $ x(t) = \cos t , y(t) = \sqrt{\sin t}$ so I got the parametric equation $\sigma (t) = (\cos t, \sqrt{\sin t}, 0) , t\in [0,2 \pi]$. Then I found $\sigma ' (t)$ and $v(\sigma (t))$ and I calculated the integral. I found $0$ as an answer. I had problems finding the border in this surface and I am not sure if I did it right. Are there another ways to find a solution? Thanks in advance!