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Stokes theorem gives ${\int }_S(\nabla \times \mathbf{F})\cdot \hat{\mathbf{n}}dS={\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{r}$ for a sufficiently smooth vector field and consistently oriented surface and boundary. Choose the hemisphere's outward normal vector; its boundary circle is then traversed counterclockwise viewed from $+z$. Parametrize it by $\mathbf{r}=(2\cos t,2\sin t,0)$, $0\le t\le 2\pi$. On this curve, $\mathbf{F}=(8\sin t,2\cos t,0)$ and $d\mathbf{r}=(-2\sin t,2\cos t,0)dt$. ThereforeAs a check, $\nabla \times \mathbf{F}=-3\hat{\mathbf{z}}$, whose flux is $-3$ times the projected disk area $4\pi$, not $-3$ times the hemisphere's surface area. Reversing orientation gives $+12\pi$.

Choose both waves as sinusoids with a common zero phase and amplitude $A$. The trigonometric sum identity givesThe slowly varying wave envelope has constant phase when $\delta x-ϵt$ is constant. Its speed is $ϵ/\delta$, and in the close-frequency limit the group velocity isIt describes the motion of a narrow-band envelope in a medium where the constituent waves retain a coherent packet interpretation; it is not an unconditional signal-speed theorem in strongly absorbing or anomalously dispersive media. The carrier phase velocity is $v_p=\omega /k$. In a nondispersive medium, $\omega =vk$ with constant $v$, so $v_g=v=v_p$.

The proposed change of variables gives $u=x-3y$, $v=2x+y$, with inverse $x=(u+3v)/7$, $y=(v-2u)/7$. The Jacobian determinant isThe two lines of slope $-2$ become $v=1,4$, while the two lines of slope $1/3$ become $u=0,7$. Thus the parallelogram becomes $0\le u\le 7$, $1\le v\le 4$. Since $v$ stays positive, the transformed integrand has no singularity there. The double integral is