\[ \newcommand{\euler}{\text{e}} \newcommand{\p}{\partial} \newcommand{\px}{\p x} \newcommand{\py}{\p y} \newcommand{\pz}{\p z} \newcommand{\pf}{\p f} \newcommand{\pu}{\p u} \newcommand{\pv}{\p v} \newcommand{\pl}{\p \boldsymbol{l}} \newcommand{\d}{\text{d}} \newcommand{\dt}{\d t} \newcommand{\dx}{\d x} \newcommand{\dy}{\d y} \newcommand{\dr}{\d r} \newcommand{\dv}{\d v} \newcommand{\dz}{\d z} \newcommand{\du}{\d u} \newcommand{\ds}{\d s} \newcommand{\dS}{\d S} \newcommand{\pp}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\dd}[2]{\d #1 \d #2} \newcommand{\dxy}{\dx\dy} \newcommand{\dyz}{\dy\dz} \newcommand{\dzx}{\dz\dx} \newcommand{\b}{\boldsymbol} \]
\[ \iint_\Sigma f(x,y,z)\dS = \iint_{D_{xy}} f(x,y,z(x,y))\sqrt{1+z_x^2+z_y^2} \dxy \]
\[ \iint_\Sigma R(x,y,z)\dxy = \iint_{D_{xy}} R[x,y,z(x,y)]\dxy \]
where the \(\Sigma\) is oriented surface using the “upper” surface as positive, the direction should be noticed during calculation.
Example. Compute the surface integral \[ \iint_\Sigma x^2\dyz + y^2\dzx + z^2\dxy \] where \(\Sigma\) is the outside of cuboid \(\Omega:\{(x,y,z)|0\le x\le a,~ 0\le y\le b,~ 0\le z\le c\}\).
We devide the \(\Sigma = \bigcup_{i=1}^6 \Sigma_i\) as the six plain surface, while the the projections of four of them in \(xOy\) are 0, hence we only need to compute the upper surface \(\Sigma_1: z=0~~(0\le x\le a, 0\le y\le b)\) and the bottom surface \(\Sigma_2: z=c~~(0\le x\le a, 0\le y\le b)\) for part \(z^2\dxy\). \[ \iint_\Sigma z^2\dxy = \iint_{\Sigma_1} c^2\dxy - \iint_{\Sigma_2} 0\dxy = abc^2 \] Similarly we have \[ \iint_\Sigma x^2\dyz = \iint_{\Sigma_3} a^2\dyz - \iint_{\Sigma_4} 0\dyz = a^2bc \] and \[ \iint_\Sigma y^2\dzx = \iint_{\Sigma_5} b^2\dzx - \iint_{\Sigma_6} 0\dyz = ab^2c \] hence \[ \iint_\Sigma x^2\dyz + y^2\dzx + z^2\dxy = abc(a+b+c) \]
Example. Compute the surface integral \(\displaystyle \iint_\Sigma xyz\dxy\), where \(\Sigma\) is the outside of sphere \(x^2+y^2+z^2=1\) in the 1st and 8nd octant.
\[ \int_{\Sigma} xyz\dxy = 2\iint_{\Sigma_1}xyz\dxy = 2\iint_D xy\sqrt{1-x^2-y^2}\dxy \]
Notice that although the integrated function \(xyz\) is the odd function respect to \(z\), the integral is double instead of 0 due to the surface integral is oriented.
Example. Denoted \(S\) as the outside of \(x^2+y^2+z^2=1\), compute \[ I = \iint_S \frac{2\dyz}{x\cos^2x} + \frac{\dzx}{\cos^2 y} - \frac{\dxy}{z\cos^2 z} \] Notice that \[ \iint_S \frac{2\dyz}{x\cos^2x} = \iint\frac{2\dxy}{z\cos^2 z} \]
\[ I = \iint_S \frac{\dxy}{z\cos^2 z} = 2\iint_{x^2+y^2\le1} \frac{\dxy}{\sqrt{1-x^2-y^2}\cos^2\sqrt{1-x^2-y^2}} \]
\[ \iint_\Sigma P\dd yz + Q\dd zx + R\dxy = \iint_\Sigma(P\cos\alpha+Q\cos\beta+R\cos\gamma)\dS \]
where the \((\cos\alpha, \cos\beta, \cos\gamma)\) is the directional cosine of the normal vector of oriented surface \(\Sigma\) at point \(\b{n}=(x,y,z)\). Denoted \(A=(P,Q,R)\), the equation can also been represented as \[ \iint_\Sigma \b A\cdot \d\b S = \iint_\Sigma \b A\cdot \b n\dS = \iint_\Sigma A_n\dS \]
where \(\d\b S=\b n\dS\) is defined as the element of oriented surface, \(A_n\) is the projection of vector \(\b A\) in vector \(\b n\).
Proof \[ \iint_\Sigma P\dd yz + Q\dd zx + R\dxy = \lim_{\lambda\rightarrow0}\sum_{i=1}^n\left[ P \right] \]
Specially, projecting into the \(xOy\) plane: \[ \iint_\Sigma P\dd yz + Q\dd zx + R\dxy = \iint [P(-z_x) + Q(-z_y) + R]\dxy \] The integral can also be projected into \(yOz\) and \(zOx\) plane by similarly.
Example. Compute the surface integral \(\displaystyle \iint_\Sigma (z^2 + x)\dyz - z\dxy\), where \(\Sigma\) is \(z=\dfrac{1}{2}(x^2+y^2)\) between \(z=0\) and \(z=2\).
\[ \b n = (\cos\alpha, \cos\beta, \cos\gamma) = \frac{1}{\sqrt{x^2+y^2+1}}(x, y, -1) \]
\[ \begin{align} \iint_\Sigma (z^2+x)\dyz &= \iint_\Sigma (z^2+x)\cos\alpha \dS \\ &= \iint_\Sigma (z^2+x)\frac{\cos\alpha}{\cos\gamma}\dxy \\ &= \iint_\Sigma (z^2+x)(-x)\dxy \\ &= -\iint_\Sigma (xz^2 + x^2)\dxy \end{align} \] Intrinsically, we convert the \(\dyz\) into \(\dxy\) by multiplying \(z_x\), that is, \[ \dyz = (-z_x)\dxy = (-x)\dxy \\ \dzx = (-z_y)\dxy = (-y)\dxy \] Similarly process can be applied for other conversion.
Gauss formula gives the relation between the triple integral in the closed region and the surface integral of the surface of the region.
\[ \iiint_\Omega \left( \pp Px + \pp Qy + \pp Rz \right)\dv = \oiint_\Sigma P\dd yz + Q\dd zx + R\dd xy = \oiint_\Sigma (P\cos\alpha + Q\cos\beta +R\cos\gamma)\dS \]
Stokes formula is the extension of Green formula. Green formula gives the relation between the integral between line integral and the double integral on the coordinates plane, while Stokes formula extended the double integral into arbitrary suface \(\Sigma\).
\[ \iint_{\Sigma} \left( \pp Ry - \pp Qz \right)\dd yz + \iint_{\Sigma} \left( \pp Pz - \pp Rx \right)\dd zx + \iint_{\Sigma} \left( \pp Qx - \pp Py \right)\dd xy = \oint_\Gamma P\dx + Q\dy + R\dz \]
The Stokes formula can also be denoted as the determinant form:
$$ \[\begin{vmatrix} \dd yz & \dd zx & \dd xy \\ \dfrac{\p}{\p x} & \dfrac{\p}{\p y} & \dfrac{\p}{\p z} \\ P & Q & R \end{vmatrix}\] = \[\begin{vmatrix} \cos\alpha & \cos\beta & \cos\gamma \\ \dfrac{\p}{\p x} & \dfrac{\p}{\p y} & \dfrac{\p}{\p z} \\ P & Q & R \end{vmatrix}\]= _P+ Q+ R $$