Need help on Stokes Theorem in surface integral Hello and how you doing today? I just came across a problem which need to use Stokes theorem.
The problems says:
Evaluate the surface integral
$$
\int_{S}\nabla\times\vec{F}\cdot{\rm d}\vec{S}
$$
where F(x,y,z)=$(y^2)i$ + $(2xy)j$+$(xz^2)k$ and S is the surface of the paraboloid $z=
x^2+y^2$ bounded by the planes $x=0,y=0$ and $z=1$ in the first quadrant pointing upward.
I got $\nabla F$ which is $(-z^2)j$
So, I stuck at here because I dont know the boundary C in order to use Stokes theorem. So could someone please help me to start?
By the way thank you very much for taking your time and consideration to help me on this problem.

 A: It's readily to see it when you draw a picture of $z=x^2+y^2$ intersected with three planes. The boundary contains three parts:


*

*$(0,0,0)\rightarrow(0,1,1)$ along $z=y^2$ for $x=0$

*$(0,1,1)\rightarrow(1,0,1)$ along $1=x^2+y^2$ for $z=1$

*$(1,0,1)\rightarrow(0,0,0)$ along $z=x^2$ for $y=0$
A: Plotting surfaces, it's pretty much clear that we will have the following figure.

Stokes theorem states that the path integral of a vector field around those dotted line is equal to the the surface vector integral of the curl of the field bound by surface. i.e.
$$ \oint_\gamma \vec F \cdot dr = \iint_\Omega \nabla \times \vec F \cdot \hat n d\sigma $$ 
First, let's calculate the left part, the path integral around along the curve going by route $OA \to AB \to BO$.
Along $OA$, the curve is $z=x^2$ and $y=0$.
$$\int_{OA} ( y^2  \hat i +  2xy  \hat j + xz^2  \hat k) \cdot dr  = \int_{OA} xz^2 \hat k \cdot (dx \hat i  + dz \hat k ) = \int_{OA} xz^2 dz $$ 
Putting $z=x^2$ and $x = 0 \to 1$, we get 
$$
\int_0^1 2x^6dx = \frac 2 7 \hspace{1cm} (1)$$
Now along AB, $z=1$, constant.
$$\int_{AB} ( y^2  \hat i +  2xy  \hat j + xz^2  \hat k) \cdot dr  = \int_{AB} ( y^2  \hat i +  2xy  \hat j + xz^2  \hat k) \cdot ( dx \hat i  + dy \hat j)$$
The circular curve can be parametrize $x = \cos(\theta)$ and $y = \sin(\theta)$, and we know that in along the curve $\theta = 0 \to \frac \pi 2 $, and that gives $dx = -\sin\theta d\theta$ and $dy = \cos\theta d\theta$, and simplifying we get
$$\int_0^{\frac \pi 2 } \left( -\sin^3 \theta  + 2 \sin\theta \cos^2 \theta   \right) d\theta  = 0 \hspace{1cm} (2)$$
Along $OB$ we proceed in the same as along $OA$ but here $x=0$ and we get 
$$\int_{BO} \vec F \cdot dr = 0 \hspace{1cm} (3)$$
Adding $(1), (2)$ and $(3)$, we get $
\displaystyle \oint_\gamma \vec F \cdot dr = \frac 2 7 $.
The right part::
The curl of the field turns out to be $- z^2 \hat j $ and the integral is 
$$ \iint_\Sigma (-z^2 \hat j)\cdot \hat n ds $$
The surface $\displaystyle z = x^2 + y^2$ can be parameterized as $S = (r \cos\theta, r\sin\theta, r^2)$  where $r =0\to 1$ and $\theta = 0 \to \frac \pi 2$ and the surface integral can be evaluated as 
$$\int_0^1 \int_0^{\frac \pi 2 } -r^4 \hat j \cdot \left( \frac{\partial S}{\partial r } \times  \frac{\partial S}{\partial \theta } \right ) dr \; d\theta  $$
The cross product can be calculated as $
-2 r^2 \cos (\theta ) \hat i -2 r^2 \sin (\theta ) \hat j + r \sin ^2(\theta )+r \cos ^2(\theta ) \hat k$
And the integral turns out as 
$$ \int_0^1\int_0^{\frac \pi 2 }  2r^6 \sin ( \theta) d\theta dr = \frac  2  7 $$
ADDED:: For plotting surfaces, just plot all surfaces and take the surface of the parabloid that is bounded by $x=0, y=0, z=1$ on first quadrant, that is the one on the first quadrant which looks like the strip above.

