Verify Stokes Theorem vector field $\vec{F} = (y, −x, xyz)$ Verify Stokes’ theorem for the vector field $$\vec{F} = (y, −x, xyz)$$ and the open surface given in cylindrical polar coordinates by $ρ + z = a$, with $z > 0$, and $a > 0$
How do I obtain the surface and bounds?
 A: First, let's state Stokes' theorem:
$$
\oint_\Gamma F \cdot d\Gamma = \iint_S \nabla \times F \cdot dS
$$
Before starting computing integrals, one should visualize the geometry of the problem. In this case, $S$ stands for a cone geometry, whose tall is $a$ and its base is centered at $(0,0,0)$, while $\Gamma$ stands for its bound, i.e. circumference whose radius is $a$ and is centered at the origin.
In this case, the easier part is the left hand side, which is immediately obtained (taking into account the relationship between $(x,y)$ and $(r,\theta)$):
$$\oint_\Gamma F \cdot d\Gamma = \int_0^{2\pi} \left( {\begin{array}{c}
   y \\
   -x \\
   xyz \\
  \end{array} } \right) \cdot \left( {\begin{array}{c}
   -\sin(\theta) \\
   \cos(\theta) \\
   0 \\
  \end{array} } \right) a d\theta = \int_0^{2\pi} \left( {\begin{array}{c}
   a\sin(\theta) \\
   -a\cos(\theta) \\
   -z a^2 \cos(\theta)\sin(\theta) \\
  \end{array} } \right) \cdot \left( {\begin{array}{c}
   -\sin(\theta) \\
   \cos(\theta) \\
   0 \\
  \end{array} } \right) a d\theta
$$
Leading to:
$$\oint_\Gamma F \cdot d\Gamma = \int_0^{2\pi} -a^2 (\cos^2(\theta)+\sin^2(\theta)) d\theta = -2\pi a^2$$
It remains to prove the right hand side. First, start computing the curl of the $F$ field:
$$\nabla \times F =  \left( {\begin{array}{c}
   xz \\
   -yz \\
   -2 \\
  \end{array} } \right) = \left( {\begin{array}{c}
   zr\cos(\theta) \\
   -zr\sin(\theta) \\
   -2 \\
  \end{array} } \right) $$
Once the curl has been computed, go on with $dS$. You can decompose in an unitary vector and its magnitude:
$$ dS = \vert dS \vert u_{dS}$$
Where by geometry:
$$\vert dS \vert = \sqrt{2} r dz d\theta$$
And:
$$u_{dS} = \left( {\begin{array}{c}
   \cos(\theta)/\sqrt{2} \\
   \sin(\theta)/\sqrt{2} \\
   1/\sqrt{2} \\
  \end{array} } \right) $$
Once we have $\nabla \times F$ and $dS$, we can compute the integral term $\nabla \times F \cdot dS$:
$$\nabla \times F \cdot dS = [zr(\cos^2(\theta)-\sin^2(\theta))-2]rdzd\theta$$
This term has to be evaluated in all $S$. We can parametrize $S$ with $\theta$ and $r$ and remembering that $r=a-z$, i.e. $S= \left\{ (r,\theta,z) \vert 0 \leq z \leq a, 0 \leq \theta < 2\pi, r = a-z\right\}$:
$$
\iint_S \nabla \times F \cdot dS = \int_0^{2\pi} \int_0^a [z(a-z)(\cos^2(\theta)-\sin^2(\theta))-2](a-z)dz d\theta
$$
As $\int_0^{2\pi} (\cos^2(\theta)-\sin^2(\theta)) d\theta = 0$:
$$
\iint_S \nabla \times F \cdot dS = \int_0^{2\pi} \int_0^a -2(a-z)dz d\theta = -2\pi a^2
$$
Obtaining a result which, of curse, is consistent with Stokes'theorem. 
