Evaluating line integral with Stokes' Theorem

Question

I have to use Stokes' Theorem to evaluate the line integral $$ \int_{\partial S} F \cdot dx $$ where $\partial S$ is the boundary of $$S =\{x^2 +y^2 = z^4,\, 0 \le z \le 3\}$$ and $$F = (xy, y, -2xz^2)^T.$$ I parameterized the boundary curve as $\gamma = (9\cos(t), 9\sin(t), 3)^T$. I defined the region bordered by $\gamma$ as $B$. After applying Stokes' Theorem, I get this surface integral: $$\int_B rot(F) \cdot n \cdot d\sigma.$$ I calculated the rotation of F as $(0, 2z^2, -x)^T$ and the normal vector n as $(0,0,1)^T$. After inserting these values the integral becomes $$\int_B 9\cos(t) \cdot d\sigma = 0.$$ Is this the correct answer or is my calculation wrong? Thank you very much!

Answer 1

I think the end result is correct but find the way it is derived a bit confusing.

Instead of integrating over the surface $S$ we can take a simpler one with the same boundary: $$B=\{z=3,x^2+y^2\le 81\}\,.$$ The unit normal at this surface is $$\mathbf{n}=\begin{pmatrix}0\\0\\1\end{pmatrix}\,.$$ The dot product of ${\rm curl}\,F$ with $\mathbf{n}$ is $-x\,.$ Therefore $$\int_B{\rm curl}\,F\,\cdot\mathbf{n}\,d\Sigma=-\int\limits_{x^2+y^2\le 81}x\,dx\,dy=-\int_{-9}^9\int\limits_{-\sqrt{81-y^2}}^{\sqrt{81-y^2}}x\,dx\,dy=0\,.$$