\[ \newcommand{\euler}{\text{e}} \newcommand{\p}{\partial} \newcommand{\px}{\p x} \newcommand{\py}{\p y} \newcommand{\pz}{\p z} \newcommand{\pf}{\p f} \newcommand{\pu}{\p u} \newcommand{\pv}{\p v} \newcommand{\pl}{\p \boldsymbol{l}} \newcommand{\d}{\text{d}} \newcommand{\dt}{\d t} \newcommand{\dx}{\d x} \newcommand{\dy}{\d y} \newcommand{\dr}{\d r} \newcommand{\dv}{\d v} \newcommand{\dz}{\d z} \newcommand{\du}{\d u} \newcommand{\ds}{\d s} \newcommand{\pp}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\dxy}{\dx\dy} \]
\[ \int_L f(x ,y)\ds = \int_\alpha^\beta f[\varphi(t), \psi(t)]\sqrt{\varphi'^2(t)+\psi(t)'^2}\dt~~~~(\alpha < \beta) \]
That is, when computing the scalar line integral, we’re required to
Specially, if we consider the \(x\) or \(y\) as the parameters, we have \[ \int_L f(x ,y)\ds = \int_a^b f(x, y(x))\sqrt{1+y'^2(x)}\dt~~~~(a<b) \] or \[ \int_L f(x ,y)\ds = \int_a^b f(y, x(y))\sqrt{x'^2(y)+1}\dt~~~~(a<b) \]
Compute \(\displaystyle\int_L \sqrt y \ds\), where \(L\) is from \((0, 0)\) to \((1,1)\) through \(y=x^2\).
\[ \int_L \sqrt y = \int_0^1 \sqrt{x^2}\sqrt{1+(x^2)'^2}\dx = \int_0^1 x\sqrt{1+4x^2}\dx = \frac{1}{12}(5\sqrt5-1) \]
Curve Integral \[ \int_L P(x, y)\dx + Q(x, y)\dy = \int_\alpha^\beta [P(\varphi(t), \psi(t))\varphi'(t) + Q(\varphi(t), \psi(t))\psi'(t)]\dt \] That is, when computing the vector line integral, we’re required to
Specially, if we consider the \(x\) or \(y\) as the parameters, we have \[ \int_L P(x, y)\dx + Q(x, y)\dy = \int_a^b [P(x, \psi(x)) + Q(x, \psi(x))\psi'(x) ]\dx \]
or \[ \int_L P(x, y)\dx + Q(x, y)\dy = \int_a^b [P(\varphi(y), y)\varphi'(y) + Q(\varphi(y), y) ]\dy \]
Compute \(\displaystyle \int_\Gamma(z-y)\dx + (x-z)\dy + (x-y)\dz\), where \[ \Gamma: \left\{\begin{array}{ll} x^2 + y^2 = 1 \\ x - y + z = 2 \end{array}\right. \]
The direction is clockwise from the perspective of \(z\)-axis.
\[ \left\{\begin{array}{ll} x = \cos t \\ y = \sin t \\ z = 2 - \cos t + \sin t \end{array}\right. ~~~~ t:2\pi\rightarrow 0 \]
\[ \int_L P\dx + Q\dy = \int_L \left( P\cos\alpha + Q\cos\beta \right)\ds \]
Types of domain: simply connected region, multiple connected region, unconnected region.
Define the direction of edge line \(L\) of region \(D\) as: the region \(D\) must always on the left of \(L\).
\[ \iint_D\left( \frac{\p Q}{\px} - \frac{\p P}{\py} \right)\dx\dy = \oint_L P\dx + Qdy \]
Proof. (1) If \(D\) is both the X-type and Y-type \[ \iint_D \pp Qx \dxy = \int_c^d\dy\int_{\psi_1(y)}^{\psi_2(y)}\pp Qx \dx \] That is, \[ \iint_D \pp Qx \dxy = \oint_L Q(x,y)\dy \] Similarly, \[ \iint_D -\pp Py \dxy = \oint_L P(x,y)\dx \] (2) If \(D\) is not the X-type nor Y-type, we could divide \(D\) into multiple regions that fits (1).
Application. Compute the area of region using the vector line integral. \[ 2\iint_D\dx\dy = \oint_L x\dy - y\dx \]
\[ A = \frac{1}{2}\oint_L x\dy - y\dx \]
Compute the area surrounded by \(x=a\cos\theta, y=b\sin\theta\). \[ A = \frac{1}{2}\oint x\dy - y\dx = \frac{1}{2}\int_0^{2\pi} (ab\cos^2\theta + ab\sin^2\theta)\d\theta = \pi ab \]
If \(L\) is the closed smooth curve, prove \[ \oint_L 2xy\dx + x^2\dy = 0 \]
Here \(P = 2xy, Q = x^2\), based on the green formula, we have \[ \oint_L 2xy\dx + x^2\dy = \iint_D (2x - 2x)\dxy = 0 \]
Compute \(\displaystyle \oint_L \frac{x\dy - y\dx}{x^2+y^2},\) where \(L\) is the continous smooth closed curve with no repeated point, which does not pass \((0, 0)\). The direction of \(L\) is counterclockwise.
Here \(P = -\dfrac{x}{x^2+y^2},~~Q = \dfrac{x}{x^2+y^2}\), hence when \(x^2+y^2\neq 0\), we have \[ \pp Qx = \frac{y^2-x^2}{(x^2+y^2)^2} = \pp Py \] if \((0,0)\notin D\), we have \[ \oint_L \frac{x\dy - y\dx}{x^2+y^2} = 0 \] if \((0,0)\in D\), we construct a small circle \(l: x^2+y^2=r^2\) with \(r>0\), use the counterclockwise as the direction of \(l\), we have \[ \oint_{L\bigcup l^-} \frac{x\dy - y\dx}{x^2+y^2} + \oint_{l} \frac{x\dy - y\dx}{x^2+y^2} = \int_0^{2\pi} \frac{r^2\cos^2\theta + r^2\sin^2\theta}{r^2}\d\theta = 2\pi \] Remark. This example reminds the condition of green function: the constructed \(P(x), Q(y)\) must have first-order continous partial derivative on the entire region \(D\), if the condition is not complyed, we can exclude those points by constructing new regions.
Compute \(\displaystyle \iint_D \euler^{-y^2}\dx\dy\), where \(D\) is the triangle with vertex \(O(0,0), A(1,1), B(0,1)\).
Let \(P=0,~ Q= x\euler^{-y^2}\), we have \[ \iint_D \euler^{-y^2}\dx\dy = \int_{OA+AB+BO} x\euler^{-y^2}\dy = \int_0^1 x\euler^{-x^2}\dx = \frac{1}{2}(1-\euler^{-1}) \]
\[ \frac{\p P}{\py} = \pp Q x \]