Without using Stokes theorem, compute surface and line integrals Let S be the surface in $R^3$ given by the part of the sphere $x^2+y^2+z^2=2$ that lies above the plane $z=1$. Let v(x,y,z) be the vector field given by $v(x,y,z)=(z,x,z)$.
a) Without using Stokes' theorem, compute $\int_{\partial S} v\cdot dr$
b) Without using Stokes' theorem, compute $\int_s (\nabla\times v)\cdot ds$
For part (a) I am not sure if I am correct, but this is what I have done:
$\int_{\partial S} v(r)\cdot dr = \int_a^b v(p(t)) \frac{dp}{dt}dt$ (1)
$p(t)=r(\sqrt2sin(q)cos(t),\sqrt2sin(q)sin(t),\sqrt2cos(q))$ from spherical coordinates
Using that the surface is above z=1, we have $\sqrt2cos(q)=1$ therefore, $cos(q)=\frac1{\sqrt2}$
This gives: $p(t) = (cos(t), sin(t),1) for 0\leq t\leq 2\pi$
$v(p(t))=(1, cos(t), 1), \frac{dp}{dt}=(-sin(t), cos(t), 0)$, plugging this into (1) we get:
$\int_{\partial S} v(r)\cdot dr = \int_0^{2\pi}-sin(t)+cos^2(t)dt = \pi$
I am not sure if I have done the part where I put the z=1 into the paramaterised equation correctly.
For part (b) I cant work out what to do when paramaterising for r  with z=1, using the equation:
$\int_S \nabla\times g\cdot ds = \int_D \nabla\times g(r(u,v))\cdot N(u,v) dudv$
 A: For part (a), note that the curve $\partial S$ is the intersection of the sphere $x^2+y^2 +z^2=2$ and $z=1$ which is simply the circle given by $x^2+y^2=1$ and $z=1$. This can be parametrized as $\mathbf{r}=(\cos t,\sin t,1)$ for $t\in[0,2\pi)$, so the integral may be written as 
$$\int_{\partial S} \mathbf{v}\cdot d\mathbf{r}=\int_{0}^{2\pi}\langle 1,\cos t,1 \rangle\cdot \langle -\sin t, \cos t,0\rangle\,dt=\int_{0}^{2\pi}(\cos^2 t-\sin t)\,dt=\pi.$$ In this equality, we have recalled that $\sin t$ has average value of zero on this interval and $\int_{0}^{2\pi} \cos^2 t\,dt=\int_{0}^{2\pi} \sin^2 t\,dt=\frac{1}{2}\cdot 2\pi =\pi$. This is the same result found in your answer.
For part (b), the region $S$ is described in spherical coordinates by $r=\sqrt{2}$ and $r \cos \theta \geq 1\implies 0\leq \theta \leq \cos^{-1}\frac{1}{\sqrt{2}}=\frac{\pi}{4}$. Now, the curl of $\mathbf{F}$ is 
$$\nabla\times F=\left|\begin{array} .\hat{x} & \hat{y} & \hat{z} \\ \partial_x & \partial_y & \partial_z \\ z & x & z\end{array}\right|=\hat{y}+\hat{z}.$$
We then want to compute the flux of this vector field through the spherical cap. We can our lives much easier by appealing to symmetry: the flux of $\hat{y}$ trhough the the portion with $y>0$ is equal and opposite to the flux through the portion with $y<0$. Consequently its contribution cancels, and we need only worry about the $\hat{z}$ component. (Let me know if you want to see this more explicitly.).  Since the area element in spherical coordinates is $d\mathbf{s}=r^2 \sin\theta \,d\theta d\phi\,\hat{r}$, the integrand of the surface integral on the sphere $r=\sqrt{2}$ is $$(\nabla\times \mathbf{v})\cdot d\mathbf{s}=2\sin\theta \, d\theta d\phi \, (\hat{z}\cdot \hat{r})=2\sin\theta \cos\theta\, d\theta d\phi$$ where we've computed the dot product by inspection of the geometry. So the integral is 
$$\int_S \mathbf{v}\cdot d\mathbf{s} = \int_0^{2\pi}\int_0^{\pi/4}\underbrace{2\sin\theta\cos\theta}_{=\sin2\theta} \,d\theta d\phi=-\pi\Big[\cos 2\theta\Big]_0^{\pi/4}=\pi.$$
These two answers match, as they must by Stokes' theorem. For a cross-check, note that Stokes' theorem also applies to the disk $D$ given by the intersection of $x^2+y^2\leq 1$ with $z=1$. The vector area element is then in the $\hat{z}$ direction, so $\mathbf{v}\cdot d\mathbf{S}=(\hat{y}+\hat{z})\cdot ds\hat{z}=ds$ and thus the surface integral is just the surface area $\int_D ds.$ Since the disk has radius $1$, this is just $\pi$ as above.
