I've been asked the following.
Verify Stokes' theorem for the vector field $\vec{a}=\vec{r}\times\hat{k}$, where $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$, and $\lbrace\hat{i},\hat{j},\hat{k}\rbrace$ is the standard basis for $\mathbb{R}^3$, given the hemisphere $C$ of radius $c$ centered at $(0,0,0)$ in the upper half region of $\mathbb{R}^3$ (that is, $z\geq0$).
My attmept is as follows. Note firstly that $\vec{a}=y\hat{i}-x\hat{j}+0\hat{k}$. Note also that, due to the statement of stokes theorem, we may be immediately interested in obtaining an expression for the curl of $\vec{a}$, which is given by $\vec{\nabla}\times\vec{a}=0\hat{i}+0\hat{j}-2\hat{k}$. From here, note that the formulation of the question suggests that $C$ is the capping surface we are to use. For this reason, note that we may take the circular contour of radius $c$ centered at $(0,0,0)$ in the plane $z=0$, oriented counter-clockwise as the boundary of $C$, which we'll of course denote $\partial C$. Notice from here that $\gamma(t)=c\cos(t)\hat{i}+c\sin(t)\hat{j}+0\hat{k}$, for $0\leq t\leq 2\pi$, parameterizes $\partial C$. For this reason, consider $$ \oint_{\partial C}\vec{a}\cdot d\vec{r}=\int_0^{2\pi}\vec{a}(\gamma(t))\cdot\gamma'(t)~dt $$ $$ =\int_0^{2\pi}(c\sin(t)\hat{i}-c\cos(t)\hat{j}+0\hat{k})\cdot(-c\sin(t)\hat{i}+c\cos(t)\hat{j}+0\hat{k})~dt $$ $$ =-c^2\int_0^{2\pi}~dt=-2c^2\pi $$ And hence, by Stokes' theorem, this quantity should equate to $$ I=\iint_C\mathrm{curl}(\vec{a})\cdot\hat{n}_C~dS $$ Where $\hat{n}_C$ denotes the unit normal to the surface of $C$ (possibly as a function of $x$, $y$, and $z$, given $(x,y,z)\in C$). To ensure the preservation of orientation, we employ the trusty right hand rule to arrive at the conclusion that the normal to $C$ is given by $\hat{n}_C=\vec{r}/|\vec{r}|$. Note, however, that on the surface of $C$, $|\vec{r}|=c$, and so $\hat{n}_C=\vec{r}/c$. For this reason, we have that $$ \mathrm{curl}(\vec{a})\cdot\hat{n}_C=(\vec{\nabla}\times\vec{a})\cdot\hat{n}_C=-\frac{2z}{c} $$ Since, on $C$, $z=\sqrt{c^2-x^2-y^2}$ (we omit consideration of the negative square root of $z^2$, since we are told that $z\geq0$), we have that $$ \mathrm{curl}(\vec{a})\cdot\hat{n}_C=-\frac{2}{c}\sqrt{c^2-x^2-y^2} $$ And, for this reason, $I$ becomes $$ I=-\frac{2}{c}\iint_C\sqrt{c^2-x^2-y^2}~dS $$ We may, from here, convert to polar coordinates to arrive at the conclusion that $$ I=-\frac{2}{c}\int_0^{2\pi}\int_0^cr\sqrt{c^2-r^2}~dr~d\theta $$ $$ =-\frac{2}{c}\int_0^{2\pi}-\frac{1}{3}\left[(c^2-r^2)^{3/2}\right]_0^c~d\theta $$ $$ =-\frac{2}{c}\int_0^{2\pi}\frac{c^3}{3}~d\theta=-\frac{4}{3}c^2\pi $$ This result, however, is obviously not equal to $-2c^2\pi$, meaning that $$ \oint_{\partial C}\vec{a}\cdot d\vec{r}\neq\iint_C\mathrm{curl}(\vec{a})\cdot\hat{n}_C~dS $$ which, as far as my understanding of Stokes' theorem is concerned, cannot be the case. Can someone perhaps help me identify the mistake that I've made?
I think your mistake is that you treated $dS$ as if it were $dx \, dy$ when you changed to polar coordinates. But $dS$ is the area element of the surface: if the surface $S$ is the graph of $z=f(x,y)$, then
$$ dS = \sqrt{1+f_x^2 + f_y^2} \, dx \, dy $$
So in your case, since the hemisphere is $z= \sqrt{c^2-x^2-y^2}$, this is
$$ dS = \frac{c}{\sqrt{c^2-x^2-y^2}} \, dx \, dy $$
This cancels with the $\sqrt{c^2-x^2-y^2}$ that's already in your integral, so you get
$$ I = -2 \iint\limits_D dx \, dy $$
where $D$ is the disc of radius $c$ in the $x,y$-plane. This is $-2$ times the area of the disc, which is $-2\pi c^2$, which agrees with the line integral.