 # Green’s Theorem

What is Green’s Theorem

Green’s Theorem gives you a relationship between the line integral of a 2D vector field over a closed path in a plane and the double integral over the region that it encloses. However, the integral of a 2D conservative field over a closed path is zero is a type of special case in Green’s Theorem.

Green’s Theorem is commonly used for the integration of lines when combined with a curved plane. It is used to integrate the derivatives in a plane. If the line integral is given, it is converted into the surface integral or the double integral or vice versa with the help of this theorem. In this article, you would learn about Green’s Theorem, what is Green’s Theorem, formula, proof, and solved examples in detail.

What is Green’s Theorem?

Green’s Theorem is amongst the four fundamental theorems of calculus wherein all the four are closely related to each other. Once you are aware of the concept of the surface integral and line integral, you would know how the Stokes theorem is based on the principle of linking the microscopic and macroscopic circulations. Similarly, the Green’s Theorem defines the relationship between the macroscopic circulation of the curve C and the sum of the microscopic circulation which is inside this curve.

Green’s Theorem Statement

Green’s Theorem states that a line integral around the boundary of the plane region D can be computed as the double integral over the region D.

Let C be a positively oriented, smooth and closed curve in a plane, and let D to be the region that is bounded by the region C. Consider P and Q to be the functions of (x, y) that are defined on the open region that contains D, and have continuous partial derivatives, then,

c(Pdx+Qdy) = ∫∫D(∂Q/∂x - ∂P/∂y)dxdy

where the path integral is traversed anti-clockwise.

Green’s Theorem Proof

You will find the Green’s Theorem proof here. Consider P (x,y) and Q (x, y) to be two smooth functions on a closed region D with a piecewise smooth boundary.

Then you would get

∂D Pdx + Qdy =  ∫∫D(∂Q/∂x - ∂P/∂y)dA,

∂D - Qdx + pdy =  ∫∫D(∂P/∂x - ∂Q/∂y)dA.

The second formula here follows the first formula in which P is replaced by -Q and Q is replaced by P. we will prove this theorem using the simplest in which the region D is a rectangle.

The line integral in 3D is a total of four different partial integrals.

∂D Pdx + Qdy = a1a2 P(x,b1)dx +  b1b2 Q(a2,y)dy + a2a1 P(x,b2)dx + b2b1 Q(a1,y)

dy  = b1b2 Q(a2,y) - Q(a1,y)dy - a2a1P(x,b2) - P(x,b1)dx.

Using the Fundamental Theorem of Calculus,

Q(a2,y) - Q(a1,y) = a1a2(∂Q/∂x)dx,

P(x,b2) - P(x,b1) = b1b2(∂P/∂y)dy

Hence, you get

∂D Pdx + Qdy = b1b2 a1a2 (∂Q/∂x)dxdy - a1a2 b1b2(∂P/∂y)dydx

= b1b2 a1a2 (∂Q/∂x) -  (∂P/∂y) dxdy

You can apply Green’s Theorem for evaluating a line integral through double integration, or for evaluating a double integral through the line integration.

Green’s Theorem Example

1. Evaluate the following integral

c (y² dx + x² dy)

where C is the boundary of the upper half of the unit desk that is traversed counterclockwise.

Solution

Since the boundary is piecewise-defined, it would be tedious to compute the integral directly.

According to Green’s Theorem,

c (y² dx + x² dy) = ∫∫D(2x-2y)dxdy

wherein D is the upper half of the disk. You can compute this integral easily now.

-11 0$\sqrt{1-x²}$ (2x - 2y) dydx = -11 (2xy - y²) $l_{0}^{\sqrt{1-x²}}$dx

= -11(2x$\sqrt{1-x²}$) - (1-x²)) dx

= 0 - -11 (1-x²) dx

= - (x - x³/3) $l_{-1}^{1}$ = -2 + ⅔ = - 4/3

Green’s Theorem Problems

1. Use Green’s Theorem to Prove the Work Determined by the Force Field

F = (x-xy)$\hat{i}$+ y²j

when a particle moves counterclockwise along the rectangle whose vertices are given as (0,0) ,  (4,0) ,  (4,6) , and  (0,6).

Solution:

Using Green’s Theorem, you find

Nₓ - Mᵧ = 0 - (-x) = x

Since the region is a rectangle, the limits are constant. Hence,

04 06 x dydx = 04 $[xy]_{0}^{6}$

= 04 6x dx

= $[3x²]_{0}^{4}$

= 48

2. Calculate the line integral of

c (3 - x) dx + (x - y) dy

in which C is the union of a unit circle which is centred at the origin and is oriented negatively, and the circle that has radius 2 and is centred at the origin and is oriented positively.

Solution:

Since the region is not simply connected, you cannot use Green’s Theorem directly here. However, if you consider the region to be the union on its left half and right half, you would see that the extra cuts would cancel each other.

You can then use Green’s Theorem on each separate piece.

Doing so you get,

Nₓ - Mᵧ = 1 - 0 = 1

The line integral would be the double integral of 1 that is the area of the region. Thus, the area is,

π(2² - 1²) = 4π