Verifying Stokes' Theorem Verify Stokes' Theorem for the given vector field $f(x, y, z)$ and surface $\Sigma$.
$$f(x, y, z) = 2y \textbf{i} - x \textbf{j} + z \textbf{k}; \quad \Sigma : x^2 + y^2 + z^2 = 1, z \ge 0$$
This was the solution given. 



I understand the line integral part, but not the surface integral. Could someone explain each of the steps? Also, the previous answers to the question said to parameterize in spherical coordinates, but this doesn't. Could someone explain that alternate solution as well?
Thank you. 
 A: Hint: The equation $x^2 + y^2 + z^2 = 1$ is the equation of the sphere in $\mathbb{R}^3$ of radius $1$ centred at the origin. The condition $z \geq 0$ gives you the upper hemisphere, and $z = 0$ corresponds to the boundary of this surface (the boundary is a curve, namely the circle $x^2 + y^2 = 1$ in the $xy$-plane). To parameterise the surface, you may want to use spherical coordinates.
A: Because your edit says that you understand the line integral part, I'll only do the surface integral.
First off, we need to consider whether Stokes' theorem actually applies here.  What is the surface?  Hopefully you recognize the formula, and can see that it's the top half of a sphere.  The boundary of this surface is then the intersection of this hemisphere with the $xy$-plane -- that is, the unit circle: $x^2 + y^2=1$.  That boundary is closed and that surface is orientable so Stokes' theorem applies.
Next, let's go ahead and find the curl of that field.  $$\mathbf f(x,y,z)=(2y, -x, z)$$ $$\nabla \times \mathbf f = \left(\frac {\partial \mathbf f_3}{\partial y}-\frac {\partial \mathbf f_2}{\partial z},\frac {\partial \mathbf f_1}{\partial z}-\frac {\partial \mathbf f_3}{\partial x}, \frac {\partial \mathbf f_2}{\partial x}-\frac {\partial \mathbf f_1}{\partial y}\right)=\left( 0,0,-3 \right)$$
Now, it's important to realize that $\mathbf n \text{dS}=(\mathbf r_\theta \times \mathbf r_\phi) \text{d}\theta\text{d}\phi$, for some parametrization of $S$: $\mathbf r(\theta, \phi)$.  So let's parametrize this surface.  While there are an infinite number of possible parametrizations of this surface, we should recognize that there is a common parametrization for spheres (and parts of spheres): $\mathbf r(\theta, \phi)=\left(\cos(\theta)\sin(\phi),\sin(\theta)\sin(\phi),\cos(\phi)\right)$.  In this case, these variables are defined over $0 \le \theta \le 2\pi$ and $\frac {-\pi}{2} \le \phi \le 0$.  If this doesn't make sense to you, check out Khan Academy's videos on surface integrals.
Notice that because $\nabla \times \mathbf f = \left(0,0,-3\right)$ is not actually dependent on $x,y,$ or $z$ in this case, we won't need to plug in expressions in $\phi$ and $\theta$ -- we can just use it as is when we compute the integral.  But in general, any $x$'s would become $\cos(\theta)\sin(\phi)$'s and similar substitution would occur for $y$'s and $z$'s if $\nabla \times \mathbf f$ weren't constant.
Now let's consider orientation.  The hemisphere does not naturally have an orientation.  But to use Stokes' theorem, we must apply one.  All we need here is to check whether the orientation we chose for the line integral is the same as that for the surface integral -- use the right hand rule.  This tells us that I wrote it right the first time: $\mathbf r_\theta \times \mathbf r_\phi$, as opposed to $\mathbf r_\phi \times \mathbf r_\theta$, but you should convince yourself of this before you move on.
Now let's compute $\mathbf r_\theta \times \mathbf r_\phi$: $$\mathbf r_\theta = \left(-\sin(\phi)\sin(\theta), \cos(\theta)\sin(\phi), 0\right)$$ $$\mathbf r_\phi = \left( \cos(\theta)\cos(\phi), \sin(\theta)\cos(\phi), -\sin(\phi) \right)$$
$$\implies \mathbf r_\theta \times \mathbf r_\phi= \left(-\sin^2(\phi)\cos(\theta), -\sin^2(\phi)\sin(\theta), -\sin(\phi)\cos(\phi) \right)$$
NOTE: If we were smart, we wouldn't even compute the first 2 components of this cross product, because we're about to dot this with the above curl, which has 0's in its first two.  I went ahead and did it here, but that's just something to keep in mind in the future.
Now, we're about ready to compute the surface integral: $$\iint_S \nabla \times \mathbf f \cdot \mathbf n \text{dS}$$ $$= \int_{\frac{-\pi}{2}}^0 \int_0^{2 \pi} \left(0,0,-3\right) \cdot \left(-\sin^2(\phi)\cos(\theta), -\sin^2(\phi)\sin(\theta), -\sin(\phi)\cos(\phi) \right) \text{d}\theta\text{d}\phi$$ $$ = \int_{\frac{-\pi}{2}}^0 \int_0^{2 \pi} 3\sin(\phi)\cos(\phi) \text{d}\theta\text{d}\phi = 6\pi \int_{\frac{-\pi}{2}}^0 \sin(\phi)\cos(\phi) \text{d}\phi =-3 \pi$$
A: Stoke's theorem states that for a oriented, smooth surface $\Sigma$ bounded simple, closed curve $C$ with positive orientation that
$$\iint_\Sigma \nabla \times F\cdot d \Sigma=\int_C F\cdot d r$$
for a vector field $F$, where $\nabla\times F$ denotes the curl of $F$.
Now the surface in question is the positive hemisphere of the unit sphere that is centered at the origin. The computation of this integral would be made much easier if you were to transform to spherical coordinates. Since we are considering the positive portion of the unit sphere centered at the origin you should integrate over $0\leq\theta\leq2\pi$ and $\frac{\pi}{2}\leq\phi\leq\pi$ using the mathematical (not the physics) convention for the angles $\theta$ and $\phi$.
As for the line integral, $C$ is the unit circle centered at the origin in the $xy$-plane. This is because we are considering the upper hemisphere of the unit sphere centered at the origin. Parametrizing this circle in a positive orientation should be easy with $\sin t$ and $\cos t$.
