Im trying to use Stokes Theorem to solve the following integral: $$\int_C(x^{2}z)dx+ (3x)dy+ (-y^3)dz$$ where $C$ is the unit circle $x^2 + y^2 = 1$, oriented counter clockwise.
This is my attempt thus far: $\textbf{F} = \langle x^2z,\:3x,\: -y^3\rangle$
$\text{Curl}\: \textbf{F} = \nabla \times \textbf{F} = $$ \begin{vmatrix} \mathrm i&\mathrm j&\mathrm k\\ \partial_x&\partial_y&\partial_z\\ x^2z&3x&-y^3 \end{vmatrix} =\langle -3y^2, x^2, 3 \rangle. $
Now, what is my d$\textbf{S} = \textbf{n}\:d\textbf{S}$? My surface is just the unit circle on the xy-plane (z=0) with radius 1 right? Since it's oriented counter clockwise, then $\textbf{n}$ is just <0,0,1>. Then, $$(\nabla \times \textbf{F}) \cdot \textbf{n}\:d\textbf{S} = \langle-3y^2, x^2,3\rangle\cdot \langle 0,0,1 \rangle = 3$$
$$ =\iint_D 3 \:dA $$
Since $dA$ is just the area of the circle with radius 1 , then we have $3\pi(1)^2 = 3\pi$.
Is this logic sound? The correct solution is $3\pi$ but I am not sure if my steps to get the answer is right. Any feedback / clarification will be nice. Thank you!