Why does Stokes theorem apply to this situation?

Question

I'm thinking Green's theorem or stokes theorem, but I don't know. It has been driving me crazy all day. Help me out here! And if you don't want to help because you know it's homework, give me some hints please :'(

$\vec{F}(x,y,z) = (2xyz + \sin(x))\vec{i} + (x^2z)\vec{j} + (x^2y)\vec{k}$

$$\int_{c} F \cdot dr$$

With parametrization given by: $c(t) = (\cos^5(t),\sin^3(t),t^4)$

I was told that Stokes theorem confirms that the integral of this is zero but I'm not understanding why Stokes theorem applies or how to know to even use it

Answer 1

Note that $$\text{curl}\left(\vec{F}\right)=\text{det}\begin{pmatrix} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2xyz +\text{sin}(x) & x^2z & x^2y\end{pmatrix}\,\,.$$ Stokes' theorem says that $$\int_C \vec{F} \cdot \text{d}\vec{r}=\int\int_S \text{curl}\left(\vec{F}\right) \cdot \text{d}\vec{S}$$ for a simple, closed boundary curve $C$ of a smooth surface $S$ (and any vector field $\vec{F}$). Observe that if $\text{curl}\left(\vec{F}\right) = \vec{0}$, then the integral is zero, regardless of the surface which $C=c(t)$ bounds. The most important part of Stokes' theorem to recognize that clarifies your confusion is that: the surface $S$ can be any surface as long as its boundary curve is given by $C$.

Answer 2

Recall Stokes theorem is the following:

$$ \int_c F \cdot dr = \int_S ( \nabla \times F) \cdot dS $$

In this case we see that

$$ \nabla \times F = \vec{0}$$

Therefore we have that

$$ \int_c F \cdot dr = 0$$