Applying Stokes Theorem Without Vector Field Q: Evaluate $\oint_S 4x dx + 9y dy + 3(x^2 +y^2) dz$ where $S$ is the boundary of the surface $z=4-x^2-y^2$ where $x,y,z \ge 0$, oriented counterclockwise as viewed from above.
I am a bit confused with how to do this question since I'm not given a vector field. Do I not apply Stoke's theorem? Any push in the right direction would be greatly appreciated.
Edit: I now know that I can express this as a vector field F $=4x$i $+9y$j $+3(x^2+y^2)$k. Using the cross product, I have found curlF$=6y$i $-6x$j, and want to evaluate $\oint_S 4x dx + 9y dy + 3(x^2 +y^2) dz$ using Stokes' theorem ($\iint_S$ curlF $\cdot$ n $dA$). However, I am unsure how to calculate n, the unit normal of $S$ which in this case would be the surface $z=4-x^2-y^2$ restricted to $x,y,z \ge 0$. Does anyone have any pointers?
 A: The Stokes theorem that you are looking for is the one using differential forms:
$$
\int_\gamma \omega = \int_S d\,\omega.
$$
Here $\gamma = \partial S$, the boundary of the surface S. (I changed your notation so that the surface is $S$ and the boundary is $\gamma$.) In this case, the surface is the portion of the parabaloid over the 1st quadrant and the boundary consists of three curves,the quarter circle $\gamma_1$ from (2,0) to (0,2) in the $xy$-plane, the portion of the parabola $\gamma_2$ in the $yz$ plane from (0,2,0) to (0,0,4), and the portion of the parabola $\gamma_3$ in the $xz$-plane back from (0,0,4) to (2,0,0). The $d$ is Cartan's exterior derivative.
$$
\begin{gathered}\omega=4x\,dx+9xy\,dy+3(x^{2}+y^{2})\,dz=4x\,dx+9xy\,dy+3(4-z)\,dz\\
d\omega=9y\,dx\wedge dy
\end{gathered}
$$
We should have
$$
\int_\gamma \omega =\int_{\gamma_1} \omega +\int_{\gamma_2} \omega+\int_{\gamma_3} \omega = \int_S d\,\omega.
$$
Let us calculate the surface integral first. The surface $S$ is parametrized by
$$
(r\cos\theta,r\sin\theta,4-r^{2})\qquad\begin{cases}
0\le r\le 2\\
0\le\theta\le\frac{\pi}{2}
\end{cases}
$$
and
$$
\begin{aligned}d\omega & =9y\,dx\wedge dy\\
 & =9r\sin\theta\,d(r\cos\theta)\wedge d(r\sin\theta)\\
 & =9r\sin\theta\,(\cos\theta\,dr-r\sin\theta\,d\theta)\wedge d(\sin\theta\,dr+r\cos\theta\,d\theta)\\
 & =9r\sin\theta\,r\,dr\wedge d\theta=9r^{2}\sin\theta\,dr\wedge d\theta.
\end{aligned}
$$
With this parameterization the surface integral becomes the simple double integral
$$
\int^{\pi/2}_0 \int^2_0 9r\sin\theta\,dr\,d\theta
=\left.\left.-3r^{3}\cos\theta\right|_{r=0}^{2}\right|_{\theta=0}^{\pi/2}=24.
$$
Of course we should to do the line integrals to see if they agree! Let us start with the integral over $\gamma_1$. This can be parametrized by
$$
\gamma_1= (2\cos\theta,2\sin\theta,0)
$$
for $0\le \theta \le \pi/2$. Now restricting $\omega = 4x\,dx+9xy\,dy+3(4-z)\,dz$ to $\gamma_1$ gives:
$$
\left.\omega\right|_{\gamma_{1}}=
-16\cos(\theta)\sin(\theta) + 72\cos^2(\theta) \sin(\theta)
$$
and
$$
\int^{\pi/2}_0 -16\cos(\theta)\sin(\theta) + 72\cos^2(\theta) \sin(\theta)
\,d\theta = 16.
$$
The curves $\gamma_2$ is parameterized by $(0,2-t,4t-t^2)$ and the curve $\gamma_3$ is given be $(t,0,4-t^2)$ for $0\le t\le 2$. The reader can check that $\int_{\gamma_2} \omega = 8$ and $\int_{\gamma_2} \omega = 0$ and find that $\int_S d\omega = \int_\gamma \omega$.
