Green's Theorem


This theorem implies that line integrals can be converted to surface integrals and vice versa. So, when a line integral is hard to calculate, the conversion to the surface integral may be considered. Let CC be a simple closed curve in xyxy-plane, and let RR be the region bounded by CC. If F1F_1 and F2F_2 are functions of (x,y)(x, y) defined on an open region containing RR and have continuous partial derivatives, then DF1dx+F2dy=R(F2xF1y)dydxline integral    surface integral\begin{aligned} \oint_D F_1 dx + F_2 dy &= \iint_R \left( \cfrac{\partial F_2}{\partial x} - \cfrac{\partial F_1}{\partial y} \right) dy dx \\ \text{line integral} &\iff \text{surface integral} \end{aligned}

Here, given a curve C:[a,b]R2C: [a, b] \in \mathbb{R}^2, “CC is simple” means that C(t0)C(t1)C(t_0) \ne C(t_1) for at0<t1<ba \leq t_0 < t_1 < b, and “CC is closed” means that the starting point C(a)C(a) and the end point C(b)C(b) are not the same. Accordingly, Green’s theorem implies that the line integral of a vector-valued function F(F1,F2)\vec{F}(F_1, F_2) is the same as the area surrounded by that line, which means here that curve.

Line Integral

While a definite integral calculates along a fixed xx-axis or yy-axis, a line integral does along an arbitrary curve which is the more generalized version. Let the parametric curve CC of the length ss be on xyxy-plane. Then, ss can be viewed as the integral of dsds in CC, and dsds is approximately a line segment. ds=(dx)2+(dy)2    s=Cds=ab(dxdt)2+(dydt)2dt\begin{aligned} ds = \sqrt{(dx)^2 + (dy)^2} \implies s = \int_C ds = \int_a^b \sqrt{\left( \cfrac{d x}{d t} \right)^2 + \left( \cfrac{d y}{d t} \right)^2} dt \end{aligned}

Therefore, if a function z=f(x,y)z = f(x, y) is integrated along CC, Cf(x,y)ds=abf(x(t),y(t))(dxdt)2+(dydt)2dt\begin{aligned} \int_C f(x, y) ds = \int_a^b f(x(t), y(t)) \sqrt{\left( \cfrac{d x}{d t} \right)^2 + \left( \cfrac{d y}{d t} \right)^2} dt \end{aligned}

Note that when f(x,y)=1f(x,y)=1, this integral is ss since this integral becomes the area of a rectangle whose width is ss and height is 11. Also, this concept can be similarly extended to nn-dimension.

Vector Field

A line integral can be also done in a vector field. A vector field is an assignment of a vector to each point in a space. When a vector-valued function F(x,y,z)\vec{F}(x, y, z) is integrated along a curve CC, this integration represents the work of F\vec{F} done by a force moving along the curve path. If the curve CC is given as X(t)=(x(t),y(t),z(t))X(t) = (x(t), y(t), z(t)), the line integral of F\vec{F} is CF(X(t))(dxdt,dydt,dzdt)dt\begin{aligned} \int_C \vec{F}(X(t)) \cdot \left( \cfrac{dx}{dt}, \cfrac{dy}{dt}, \cfrac{dz}{dt} \right) dt \end{aligned}

This can be written more simply as CFds\displaystyle \int_C \vec{F} \cdot d\vec{s}.

Discontinuous Integrand


If the integrand contains a discontinuous point, the line integral except for this point in CC becomes 00. Let C1C_1 and C2C_2 be piecewise smooth, not intersecting each other, simple closed curves as above. Then, the line integral except for the discontinuous point is CFds=C1Fds+C2Fds=0    C1Fdsundefinedcounterclockwise=C2Fdsundefinedclockwise\begin{aligned} \oint_C \vec{F} \cdot d\vec{s} = \oint_{C_1} \vec{F} \cdot d\vec{s} + \oint_{C_2} \vec{F} \cdot d\vec{s} = 0 \implies \underbrace{\oint_{C_1} \vec{F} \cdot d\vec{s}}_{\text{counterclockwise}} = - \underbrace{\oint_{C_2} \vec{F} \cdot d\vec{s}}_{\text{clockwise}} \end{aligned}

Green’s theorem can be more easily applied if a region containing the discontinuous point is considered as a circle with a radius 11( or any positive real number).


Example 1. Let CC be the circumference of the circle with the radius 11. Calculate the line integral

C(3x3+y)dx+(5x+y5)dy\begin{aligned} \oint_C (3x^3 + y)dx + (5x + y^5)dy \end{aligned}

Let F1=3x3+yF_1 = 3x^3 + y and F2=5x+y5F_2 = 5x + y^5. Then, for the region RR surrounded by CC, C(3x3+y)dx+(5x+y5)dy=R(F2xF1y)dydx=R(51)dxdy=4π\begin{aligned} \oint_C (3x^3 + y)dx + (5x + y^5)dy &= \iint_R \left( \cfrac{\partial F_2}{\partial x} - \cfrac{\partial F_1}{\partial y} \right) dy dx = \iint_R (5 - 1)dx dy = 4\pi \end{aligned}

Example 2. For C:x22+y23=1C: \cfrac{x^2}{2} + \cfrac{y^2}{3} = 1, calculate the line integral

Cydx+xdyx2+y2\begin{aligned} \oint_C \cfrac{-y dx + x dy}{x^2 + y^2} \end{aligned}

Note that the integrad has the discontinuous point at (0,0)(0, 0) which is what CC contains as well. Let C1C_1 and C2C_2 be piecewise smooth, not intersecting each other, simple closed curves. Considering C2C_2 containing (0,0)(0, 0) as a circle with a radius 11, C2C_2 is given as X(t)=(cost,sint)X(t) = (\cos t, \sin t). Cydx+xdyx2+y2=C2Fds=C2F(X(t))(dxdt,dydt)dt=02π(sint,cost)(sint,cost)dt=2π\begin{aligned} \oint_C \cfrac{-y dx + x dy}{x^2 + y^2} &= -\oint_{C_2} \vec{F} \cdot d\vec{s} = -\int_{C_2} \vec{F}(X(t)) \cdot \left( \cfrac{dx}{dt}, \cfrac{dy}{dt} \right) dt \\ &= - \int_0^{-2\pi} (- \sin t, \cos t) \cdot (-\sin t, \cos t) dt = 2\pi \end{aligned}

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