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I was trying the solve the following exercise from Apostol Vol 2: Using Stokes theorem, proof that $\int_{C} (y-z)dx +(z-x)dy+(x-y)dz = 2 {\pi}a(a+b)$, where $C$ is the intersection of the cylinder $x^{2}+y^{2}=a^{2}$ and the plane $\frac{x}{a}+\frac{z}{b}=1$, $0<a<b$. First I computed curl$(F)=(-2,-2,-2)$, then I got $S$ to be the elliptical region bounded by $C$, i.e. the region inside $\frac{{(z-b)}^{2}} {b^{2}}+\frac{y^{2}}{a^2}=1$ on the $yz$ plane. Then I computed the unitary normal vector to $S$, $n=\frac{\frac{1}{a}i+\frac{1}{b}k}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{b}{\sqrt{a^2+b^2}}i+\frac{a}{\sqrt{a^2+b^2}}k$, and got curl$(F) \cdot n=\frac{-2(a+b)}{\sqrt{a^2+b^2}}$.

Stoke's theorem states that $\int_{C}F\cdot d\alpha=\iint_S curl(F) \cdot dS$, so in this case that would be $\int_{C}F\cdot d\alpha= \frac{-2(a+b)}{\sqrt{a^2+b^2}} \iint_S dS$. The double integral in the latter formula is the area of the ellipse, $\pi ab$. The answer would be $\frac{-2ab\pi(a+b)}{\sqrt{a^2+b^2}}$, which does not resembles at all to the answer I should be getting.

I already got the right answer by taking the parametrization of the surface $(rcos(\theta), rsin(\theta),b-\frac{brcos(\theta)}{a})$, with $0\le \theta \le 2\pi$ and $0\le r \le a$ but I'm still wondering what went wrong with the steps AND reasoning I showed previously. I know that the same question was already asked here but the answers don't mention anything like this. Thanks in advance.

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    $\begingroup$ I believe one mistake you are making is your calculation for the area of the ellipse as $\pi ab$. The ellipse does not lie simply in the $(y,z)$ plane. If it did, the normal vector would be purely along $i$ after all. $\endgroup$ – Kieran Cooney Jan 15 '18 at 15:41
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    $\begingroup$ Aside from getting the area of the ellipse wrong, per the preceding comment, the sign difference between your answer and the correct one suggests that you’re using the wrong normal. $\endgroup$ – amd Jan 15 '18 at 18:14
  • $\begingroup$ For the normal, I know I should use the other one, the thing that was bothering me the most was the number that I got, thanks! If I wanted to calculate the area of that ellipse, what should I do? Or is it just not worth it (given that I already got the right thing with other method) Thanks again! $\endgroup$ – V.E.O. Jan 16 '18 at 2:28

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