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Differentiation of $x=\rho \cos \varphi$, $y=\rho \sin \varphi$ givesThe two radial face areas are ${\rho }_0\Delta \varphi \Delta z$ and $({\rho }_0+\Delta \rho )\Delta \varphi \Delta z$. Each constant-$\varphi$ face has area $\Delta \rho \Delta z$, and each constant-$z$ face has areaThe volume is this last expression times $\Delta z$.

Only the radial faces contribute to the flux of $F(\rho )e_{\rho }$. Dividing their difference by the volume and taking $\Delta \rho \to 0$ givesat $\rho ={\rho }_0$.

A field is irrotational when $\nabla \times F=0$. In ${\mathbb{R}}^3$ definewhere $\gamma$ is any path from $x_0$ to $x$. The difference between two choices is the circulation around a closed loop. By the Stokes theorem it is the flux of $\nabla \times F$ through a spanning surface and hence zero. Thus $V$ is well-defined, $V(x_0)=0$, and differentiation gives $F=-\nabla V$.

If $\nabla \times A_0=B$, every solution isfor an arbitrary differentiable scalar $\chi$, since the difference is curl-free. A necessary condition is $\nabla \cdot B=0$, because the divergence of a curl vanishes.

Existence follows from the three-transitivity of Möbius transformations: one can send three distinct source points to $0,1,\infty$ and then send those to the three distinct target points.

For uniqueness, suppose $S$ and $T$ have the same values at three distinct points. Then $M=T^{-1}S$ fixes those points. Ifthe finite fixed points satisfy $c{z}^2+(d-a)z-b=0$. A nonidentity Möbius map therefore has at most two fixed points on the Riemann sphere. Since $M$ has three, it is the identity, and $S=T$.

Expanding with the Levi-Civita symbol and the product rule givesThe Stokes theorem stateswhere the boundary orientation follows the right-hand rule about $n$.

Apply it to $F=m\times v$. On $S$, $m=n$ and $v\cdot n=0$. Substitution of the vector identity and contraction with $n$ leavesThe boundary integrand satisfies $(n\times v)\cdot dx=v\cdot (dx\times n)$, proving the formula. The vector $dx\times n$ is the outward co-normal in the tangent plane of $S$.

For the hemisphere, $n=e_r$ and $v=r\sin \theta e_{\theta }$. The projected divergence is $2\cos \theta$, so the left side isOn the positively oriented equator, $dx=R{e}_{\varphi }d\varphi$, hence $dx\times n=R{e}_{\theta }d\varphi$ and the boundary integral is also $2\pi R^2$.

The parameter sensitivity equation follows by differentiating with respect to $\mu$ and writing $u=y_{\mu }$:At $\mu =0$, $y_x=y$, $y(0)=1$, so $y(x,0)=e^x$. Thenand the integrating factor $e^{-x}$ yields

Equating powers in $y=y_0+\mu y_1+{\mu }^2{y}_2+\cdots$ givesConsequently