I am currently undertaking a class on Classical Electrodynamics, and I am realising my multivariable calculus skills are slightly rusty, as it has been a few years.
Anyhow, I am given a surface $$S\{(x,y,z)\in\mathbb{R}^3|z=1-x^2-y^2, z>0\}$$ and a vector field $$\textbf{F}(x,y,z)=(z,x,y)$$ Which I am then tasked to sketch the surface (which is fair enough, it is obviously a paraboloid with a global maximum at $z=1$)
I am then to verify Stokes' theorem by computing both the right and left-hand sides of it manually, and compare, i.e. to compute and compare $$\oint_{\partial C}\textbf{F}\cdot \textrm{d}\textrm{S} \ \ \ \ \textrm{and} \ \ \ \ \iint_{C}(\nabla \times \textbf{F})\cdot \textrm{d}\textbf{S}$$ where $\partial C$ is the boundary of $C$
So anyway, I go about my business and parameterize the curve $$\textbf{r}(\theta)=(\cos(\theta),\sin(\theta),0), \ \ \ 0\leq\theta\leq2\pi$$ as it is the circle $x^2+y^2=1$ projected down into the $xy$-plane. By extension we then have $$\int_{\partial C}\textbf{F}(\textbf{r}(\theta))\cdot\textbf{r}'(\theta)=\int_{0}^{2\pi}\cos^2(\theta)=\pi $$
Now when I go to compute the right hand side I start by parameterizing the surface as
$$\textbf{A}(r,\theta)=(r\cos(\theta),r\sin(\theta),1-r^2), \ \ \ 0\leq \theta\leq2\pi, \ \ \ 0\leq r\leq 1$$
To compute the integral I need my $\textrm{d}\textbf{S}$, which I compute by utilizing the fundamental vector product, i.e. $$\textrm{d}\textbf{S}=\frac{\partial\textbf{A}}{\partial{r}}\times \frac{\partial \textbf{A}}{\partial\theta }=(2r^2\cos\theta,2r^2\sin\theta,r)$$ Using Curl in Cylindrical coordinates we find $$\nabla \times \textbf{F}=\nabla\times(z,x,y)=\nabla\times(z=1-r^2,r\cos\theta,r\sin\theta)=(\cos(\theta),-\sin(\theta),2\cos(\theta))$$
Edit: For some further clarification on the curl I used that
$$\nabla \times \textbf{F}=\hat{\textbf{r}}\left(\frac{1}{r}\partial_\theta F_z-\partial_zF_\theta\right)+\hat{\theta}\left(\partial_zF_r-\partial_rF_z\right)+\hat{\textbf{z}}\left(\frac{1}{r}\partial_r rF_\theta-\frac{1}{r}\partial_\theta F_r\right)$$
I first tried it with a parameterized $z=1-r^2$ which yielded the above answer, then I tried it with $z$ for the heck of it and it's identical with the only difference being the $\phi$-component being $1-\sin(\theta)$
When the first try is dotted with $\textrm{d}\textbf{S}$ it gives $$(\nabla\times\textbf{F})\cdot \textrm{d}\textbf{S}=2r^2\cos(2\theta)+2r\cos(\theta)$$
When the second try is dotted with $\textrm{d}\textbf{S}$ it gives $$-2r^2\sin^2(\theta)-2r\sin(\theta)+2r^2\cos^2(\theta)+2r\cos(\theta)=2r^2\cos(2\theta)+2r(\cos\theta-\sin\theta)$$
But when I integrate that entire shebang (well both of them) over $0$ to 2$\pi$ and $0$ to $1$ (I now realised $rdrd\theta$ was baked into dS ) it spits out zero which is not equal to one, so I am at a loss of where it is I am failing. If someone could help a poor soul out, that would be nice.