스토크스 정리

The given surface is an ellipsoid centred at $(x,y,z)=(0,0,1)$. It caps a curve $\mathcal{C}$ in the plane $z=0$, given by $x^2+y^2=4$. This is a circle of radius $2$ centred at the origin, oriented counterclockwise when viewed from the positive $z$-axis.

Method I — double Stokes': Let $\mathcal{D}$ denote the plane disk $x^2+y^2\le 4$, $z=0$. Using Stokes' theorem twice gives $$\iint _\mathcal{S} \mathbf{G}\cdot \hat{\mathbf{n}}\,\mathrm{d}S = \iint _\mathcal{S} \nabla\times\mathbf{F}\cdot \hat{\mathbf{n}}\,\mathrm{d}S = \oint_\mathcal{C} \mathbf{F}\cdot \mathrm{d}\mathbf{r} = \iint _\mathcal{D} \nabla\times\mathbf{F}\cdot \hat{\mathbf{n}}\,\mathrm{d}S = \iint _\mathcal{D} \mathbf{G}\cdot \hat{\mathbf{n}}\,\mathrm{d}S$$ Now in $\mathcal{D}$ we have $\hat{\mathbf{n}}=\hat{\mathbf{k}}$ and $z=0$, so on this surface,

$$\begin{aligned} \mathbf{G}\cdot\hat{\mathbf{n}} = (\nabla\times\mathbf{F})\cdot\hat{\mathbf{k}} &= \det\left[\begin{matrix} 0 & 0 & 1 \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ (xz - y^3\cos z) & x^3 e^z & xyze^{x^2+y^2+z^2} \end{matrix}\right]_{z=0} \\[0.05in] &= \big[3x^2 e^z + 3y^2\cos z\big]_{z=0} = 3(x^2+y^2) \end{aligned}$$

Hence, using polar coordinates, $$\iint _\mathcal{S} \mathbf{G}\cdot \hat{\mathbf{n}}\,\mathrm{d}S = \iint _\mathcal{D} 3(x^2+y^2)\,\mathrm{d}x\mathrm{d}y = 3\int_{\theta=0}^{2\pi} \int_{r=0}^2 (r^2)\,r\,\mathrm{d}r\,\mathrm{d}\theta = 3(2\pi)(4)= 24\pi$$

Method II — single Stokes': By Stokes' theorem

$$\iint _\mathcal{S} \mathbf{G}\cdot \hat{\mathbf{n}}\,\mathrm{d}S = \iint _\mathcal{S} \nabla\times\mathbf{F}\cdot \hat{\mathbf{n}}\,\mathrm{d}S = \oint_\mathcal{C} \mathbf{F}\cdot \mathrm{d}\mathbf{r}$$

Parametrize the circle $\mathcal{C}$ using $$\mathbf{r}(\theta) = 2\cos\theta\,\hat{\boldsymbol{\imath}} + 2\sin\theta\,\hat{\boldsymbol{\jmath}},\qquad 0\le\theta\le 2\pi$$ to obtain $$\mathrm{d}\mathbf{r} = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}\theta}\,\mathrm{d}\theta = (-2\sin\theta\,\hat{\boldsymbol{\imath}} + 2\cos\theta\,\hat{\boldsymbol{\jmath}})\,\mathrm{d}\theta.$$ Then since $z=0$ on $\mathcal{C}$,

$$\begin{aligned} \iint _\mathcal{S} \mathbf{G}\cdot \hat{\mathbf{n}}\,\mathrm{d}S &= \int_{0}^{2\pi}\big( \overbrace{-(2\sin\theta)^3\,\hat{\boldsymbol{\imath}} + (2\cos\theta)^3\,\hat{\boldsymbol{\jmath}}}^ {\mathbf{F}\big(\mathbf{r}(\theta)\big)}\big) \cdot \big(\overbrace{-2\sin\theta\,\hat{\boldsymbol{\imath}} + 2\cos\theta\,\hat{\boldsymbol{\jmath}}}^{\mathbf{r}'(\theta)}\big) \,\mathrm{d}\theta \\ &= 16 \int_{0}^{2\pi}\big(\sin^4\theta + \cos^4\theta\big)\,\mathrm{d}\theta \end{aligned}$$

By the double angle trig identities

$$\begin{aligned} \cos^2\theta=\frac{1+\cos(2\theta)}{2}\qquad \sin^2\theta=\frac{1-\cos(2\theta)}{2}\qquad \end{aligned}$$

we have

$$\begin{aligned} \sin^4\theta + \cos^4\theta &=\frac{\big[1-\cos(2\theta)\big]^2}{4} + \frac{\big[1+\cos(2\theta)\big]^2}{4} \\ &=\frac{1}{2} + \frac{\cos^2(2\theta)}{2} =\frac{1}{2} + \frac{1+\cos(4\theta)}{4} \end{aligned}$$

So $$\iint _\mathcal{S} \mathbf{G}\cdot \hat{\mathbf{n}}\,\mathrm{d}S =16\int_{0}^{2\pi} \left(\frac{3}{4}+\frac{1}{4}\cos(4\theta)\right) \,\mathrm{d}\theta = 16\times\frac{3}{4}\times (2\pi) = 24\pi$$