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 C be a simple closed curve in xy-plane, and let R be the region bounded by C. If F1 and F2 are functions of (x,y) defined on an open region containing R and have continuous partial derivatives, then ∮DF1dx+F2dyline integral=∬R(∂x∂F2−∂y∂F1)dydx⟺surface integral
Here, given a curve C:[a,b]∈R2, “C is simple” means that C(t0)=C(t1) for a≤t0<t1<b, and “C is closed” means that the starting point C(a) and the end point C(b) are not the same. Accordingly, Green’s theorem implies that the line integral of a vector-valued function F(F1,F2) 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 x-axis or y-axis, a line integral does along an arbitrary curve which is the more generalized version. Let the parametric curve C of the length s be on xy-plane. Then, s can be viewed as the integral of ds in C, and ds is approximately a line segment. ds=(dx)2+(dy)2⟹s=∫Cds=∫ab(dtdx)2+(dtdy)2dt
Therefore, if a function z=f(x,y) is integrated along C, ∫Cf(x,y)ds=∫abf(x(t),y(t))(dtdx)2+(dtdy)2dt
Note that when f(x,y)=1, this integral is s since this integral becomes the area of a rectangle whose width is s and height is 1. Also, this concept can be similarly extended to n-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) is integrated along a curve C, this integration represents the work of F done by a force moving along the curve path. If the curve C is given as X(t)=(x(t),y(t),z(t)), the line integral of F is ∫CF(X(t))⋅(dtdx,dtdy,dtdz)dt
This can be written more simply as ∫CF⋅ds.
Discontinuous Integrand
If the integrand contains a discontinuous point, the line integral except for this point in C becomes 0. Let C1 and C2 be piecewise smooth, not intersecting each other, simple closed curves as above. Then, the line integral except for the discontinuous point is ∮CF⋅ds=∮C1F⋅ds+∮C2F⋅ds=0⟹counterclockwise∮C1F⋅ds=−clockwise∮C2F⋅ds
Green’s theorem can be more easily applied if a region containing the discontinuous point is considered as a circle with a radius 1( or any positive real number).
Application
Example 1. Let C be the circumference of the circle with the radius 1. Calculate the line integral
∮C(3x3+y)dx+(5x+y5)dy
Let F1=3x3+y and F2=5x+y5. Then, for the region R surrounded by C, ∮C(3x3+y)dx+(5x+y5)dy=∬R(∂x∂F2−∂y∂F1)dydx=∬R(5−1)dxdy=4π
Example 2. For C:2x2+3y2=1, calculate the line integral
∮Cx2+y2−ydx+xdy
Note that the integrad has the discontinuous point at (0,0) which is what C contains as well. Let C1 and C2 be piecewise smooth, not intersecting each other, simple closed curves. Considering C2 containing (0,0) as a circle with a radius 1, C2 is given as X(t)=(cost,sint). ∮Cx2+y2−ydx+xdy=−∮C2F⋅ds=−∫C2F(X(t))⋅(dtdx,dtdy)dt=−∫0−2π(−sint,cost)⋅(−sint,cost)dt=2π
Keep going!Keep going ×2!Give me more!Thank you, thank youFar too kind!Never gonna give me up?Never gonna let me down?Turn around and desert me!You're an addict!Son of a clapper!No wayGo back to work!This is getting out of handUnbelievablePREPOSTEROUSI N S A N I T YFEED ME A STRAY CAT