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 on surface integral. The question uses the normal unit vector instead of just the normal vector , don’t understand why.

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.

- 1$\begingroup$ 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 $\endgroup$– danimalApr 27, 2015 at 12:45

## 1 Answer

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.

- $\begingroup$ If the normal vectors cancel out. Why can’t you use it in this case? $\endgroup$– JohnApr 27, 2015 at 16:18
- $\begingroup$ I edited my answer, correcting some typos, but I also explained better my point. $\endgroup$ Apr 27, 2015 at 16:36
- $\begingroup$ 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. $\endgroup$– JohnApr 27, 2015 at 16:45
- $\begingroup$ Not correct in which sense? Do you get just a wrong sign or eveything? $\endgroup$ Apr 27, 2015 at 17:46
- $\begingroup$ Huh, that solution is wrong: the area of the circle is $4\pi^2$, clearly not $4\pi$. $\endgroup$ Apr 27, 2015 at 17:52