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Let $\mathbf{F}=3y \ \mathbf{i} -3xz \ \mathbf{j} + (x^2-y^2) \ \mathbf{k}.$ Compute the flux of the vectorfield $\text{curl}(\mathbf{F})$ through the semi-sphere $x^2+y^2+z^2=4, \ z\geq 0$, by using direct parameterization of the surface and computation of $\text{curl}(\mathbf{F}).$


Denote the semisphere by $S$. In the book they start off by noting that

$$\iint_S xy \ dS = \iint_S xz \ dS = \iint_S yz \ dS = 0 \quad (1)\\ \iint_S x^2 \ dS = \iint_S y^2 \ dS = \iint_S z^2 \ dS \quad \quad \quad (2)\\ $$

They then proceed computing the curl of the field, which is $(2x-2y,-2x,-2z-3)$ and the normal of the surface, out from the surface is $\mathbf{N}=\frac{1}{2}(x,y,z).$ So

$$\text{curl}(\mathbf{F})\cdot \mathbf{N} = \frac{1}{2}(2x^2-4xy-2z^2-3z).$$

Using the symmetries (1) and (2) we get the integral

$$\frac{1}{2}\iint_S-3z \ dS.$$

They then proceed to compute $dS$ by arguing that: $S$ is a part of the implicitly defined surface $F(x,y,z)=4$, where $F(x,y,z)=x^2+y^2+z^2,$ so

$$dS=\left|\frac{\nabla F}{F_z}\right|dxdy = \left|\frac{(2x,2y,2z)}{2z}\right| \ dxdy= \left|\frac{(x,y,z)}{z}\right| \ dxdy=\frac{2}{z} \ dxdy \quad (3).$$

Question: Can someone intuitively, and with other examples, explain the symmetry properties (1), (2) and why equation (3) holds and what it says?

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  • $\begingroup$ for (3) what is not clear to you exactly, the set up or some detail? $\endgroup$ – gimusi Mar 8 '18 at 11:13
  • $\begingroup$ This one was finally clear? it is a purely geometric fact $\endgroup$ – gimusi Mar 9 '18 at 15:59
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Note that

  • $(1)$ hold since $xy,xz,yz$ are antisymmetric function on the domain, indeed, for example, at each point xy in the I quadrant correspond -xy in th II and so on

  • $(2)$ hold since $x^2,y^2,z^2$ are symmetric function on the domain and each single integral for $x^2,y^2,z^2$ must assume the same value for every semisphere

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  • $\begingroup$ Ok, I understand. Thanks. As for (3), the entire setup. What does that equation have to do with "implicitly defined"? $\endgroup$ – Parseval Mar 8 '18 at 11:21
  • $\begingroup$ indeed it is a bit strange also for me at a first sight, but it should be equal to $$\left\lVert \vec r_x \times \vec r_y \right\rVert dxdy$$ $\endgroup$ – gimusi Mar 8 '18 at 11:33
  • $\begingroup$ ah ok the following holds $$\frac{dS}{dxdy}=\left|\frac{\nabla F}{F_z}\right|=\cos \theta$$ $\endgroup$ – gimusi Mar 8 '18 at 11:38

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