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The following problem would be typically solved applying Stokes Theorem, or even directly calculating each integral, but this may need a change of perspective. I appreciate some hints because I can not see how this could be related with the divergence theorem.

Given the field $\mathbf{F}(x,y,z)=(-y, -yz, \frac{x^2}{2})$ and the sufaces $S_1=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=4, x\ge0\}$, $S_2=\{(x,y,z)\in\mathbb{R}^3:y^2+z^2\le 4, x=0\}$ use the Divergence theorem to justify the following equation: $$ \iint_{S_1} \operatorname{curl}\mathbf{F} \cdot dS = \iint_{S_2} \operatorname{curl}\mathbf{F} \cdot dS$$

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  • $\begingroup$ When you write \text{curl}\mathbf{F} you see $\text{curl}\mathbf{F}$, but when you write \operatorname{curl}\mathbf{F} you see $\operatorname{curl}\mathbf{F}$. The later form puts spaces to the left and right or "curl" in expressions like $a\operatorname{curl} \mathbf{F}$ but if you write $a\operatorname{curl}(\mathbf{F})$ then it doesn't put the same space between "curl" and the left parenthesis. In other words, all the appropriate spacing conventions are in the software already when \operatorname{} is used. (I edited the question accordingly.) ${}\qquad{}$ $\endgroup$ Sep 15, 2015 at 19:59

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The two surfaces enclose an hemisphere $E= \{ (x,y,z ) : x^2+y^2+z^2 \le 4 , x\ge 0\}$, in particular $S_1 \cup S_2 = \partial E$. So you can use divergence theorem, recalling that $$\operatorname{div}(\operatorname{curl} \mathbf{F}) = 0$$.

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  • $\begingroup$ Right. But I don't see how this fact can be used to prove the given equality. $\endgroup$ Sep 15, 2015 at 19:40
  • $\begingroup$ You have $\int_E \operatorname{div} \operatorname{curl} F = 0$. Now, apply divergence theorem. $\endgroup$
    – Crostul
    Sep 15, 2015 at 19:44

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