Problem:
Let $w = (x + y)dz + (y + z)dx + (x + z)dy$ and let $S$ be the upper part of the unit sphere; that is, $S$ is the set of $(x,y,z)$ with $x^2+y^2+z^2 =1$ and $z\ge0$. $\delta$$S$ is the unit circle in the $xy$ plane. Evaluate $\int_{\delta S} w$ both directly and by Stokes' theorem.
Attempt: So to evaluate directly, I evaluated ${dw}$ and got $dx$^$dz$ + $dy$^$dz$ + $dy$^$dx$ + $dz$^$dx$ + $dx$^$dy$ + $dz$^$dy$ which I'm pretty sure equals 0. Therefore $\int_{\delta S} w$ = $\int_S dw$ = 0.
But how do I evaluate this using Stokes' theorem: $\int\int_S \Delta\times F \cdot dS = \int_{\delta S} F \cdot ds $?
Also is what I did above correct?
So, to evaluate $\int_{\partial S} w$ using Stokes Theorem, it is enough to compute that $\mathrm{d} w = 0$, and so on $S$ $$ \int_{\partial S} w = \int_S \mathrm{d} w = 0. $$ Fairly straightforward. For the second part, evaluating the integral directly, we find that $$ \int_{\partial S} w = \int_{x^2 + y^2 = 1} (x + y)\mathrm{d}z + (y + z)\mathrm{d}x + (x + z)\mathrm{d}y . $$ Noting that on $\partial S$ $z = \mathrm{d} z = 0$, we can simplify this to the form $$ \int_{x^2 + y^2 = 1} y \mathrm{d} x + x \mathrm{d} y = \int_{0} ^{2 \pi} \sin \theta \cdot (- \sin \theta ) + \cos \theta \cdot (\cos \theta ) \, \mathrm{d} \theta . $$ This integral of course can be computed - however, lets think about this for a moment: cosine and sine are periodically equivalent in the sense that $\sin \theta = \cos (\theta - \frac{\pi}{2})$. So it is obvious and straightforward that $$ \int_{0} ^{2 \pi} \sin ^2 \theta \, \mathrm{d} \theta = \int_{0} ^{2 \pi} \cos ^2 \theta \, \mathrm{d} \theta $$ from which it follows that $$ \int_{\partial S} w = 0. $$