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As the title states. I do not know why the solution used the normal unit vector. I would just use r(u,v) = ui + vj + (-u-v)k and ru x rv = i+j+k to get the result. The question and the solution are attached.Question

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John
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    if you used a non-unit normal vector then you would get the wrong result from Stokes' theorem (you would be out by a multiplicative constant). Using the unit normal means you can separate out the direction and magnitude of the elemental area vector – danimal Apr 27 '15 at 12:45

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That normal unit vector is always the right way to go. That said, let's start from the surface integral $$ \int_S\operatorname{curl} F\cdot\hat{n}\ \mathrm{d}\sigma $$ where $S$ is the surface, $\hat{n}$ is the unit normal vector and $\mathrm{d} \sigma$ the "surface measure".

Let's call $\varphi=\varphi(u,v)$ a function that maps a set $D\subset\mathbb R^2$ onto the surface $S$. Using this parameterization of $S$, you find the normal vector from $\frac{\partial\varphi}{\partial u}(u,v)\times\frac{\partial\varphi}{\partial v}(u,v)$. The normal unit vector will be $$ \hat{n}(x,y,z)=\hat{n}\big(\varphi(u,v)\big)=\frac{\varphi_u(u,v)\times\varphi_v(u,v)}{\|\varphi_u(u,v)\times\varphi_v(u,v)\|} $$ but if you use the same parameterization to transform the surface integral to a 2d integral you get $$ \int_D\operatorname{curl}F\big(\varphi(u,v)\big)\cdot\frac{\varphi_u(u,v)\times\varphi_v(u,v)}{\|\varphi_u(u,v)\times\varphi_v(u,v)\|} \|\varphi_u(u,v)\times\varphi_v(u,v)\|\ \mathrm{d}u\ \mathrm{d}v $$

In the end, in this cases it doesn't really matter that the normal vector (as long as the parameterization is the same!) has norm 1, you get the same result.

yellon
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  • If the normal vectors cancel out. Why can't you use it in this case? – John Apr 27 '15 at 16:18
  • I edited my answer, correcting some typos, but I also explained better my point. – yellon Apr 27 '15 at 16:36
  • I understand what you're trying to say about the unit normal vector cancelling itself out. Which means that any normal vector would do. But when I use the normal vector i+j+k , the answer is not correct. – John Apr 27 '15 at 16:45
  • Not correct in which sense? Do you get just a wrong sign or eveything? – yellon Apr 27 '15 at 17:46
  • Huh, that solution is wrong: the area of the circle is $4\pi^2$, clearly not $4\pi$. – yellon Apr 27 '15 at 17:52
  • The area of the circle should be 4pi , 22pi = 4pi. and my answer is completely off. – John Apr 27 '15 at 18:45
  • Omg you're right. I'll go hide from the world for some time. – yellon Apr 27 '15 at 19:04
  • I tried calculating that integral and I find $\sqrt{3}$ times the area of the circle too. Maybe you used the normal vector you found as the solution says? That would be wrong of course, that works only when you're parameterizing the surface of course. – yellon Apr 27 '15 at 19:17