# Green’s Theorem

An important theorem of integration is “Green’s Theorem”. Green’s theorem is mainly used for the integration of line combined with a curved plane. Green’s theorem shows the relationship between a line integral and a surface integral. It is related to many theorems such as Gauss theorem, Stokes theorem. This theorem is used to integrate the derivatives in a particular plane. If a line integral is given, it is converted into surface integral or the double integral or vice versa using this theorem.

## Green’s Theorem Statement

Let C be the positively oriented, smooth, and simple closed curve in a plane, and D be the region bounded by the C. If L and M are the functions of (x, y) defined on the open region, containing D and have continuous partial derivatives, then the Green’s theorem is stated as

$\oint_{C}(Ldx+Mdy)= \iint_{D}(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial x})dxdy$

Where the path integral is traversed counterclockwise.

## Green’s Theorem Proof

To prove: $\oint_{C}(Ldx+Mdy)= \iint_{D}(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial x})dxdy$

Proof:

From the given diagram, we get

$\oint_{c}Ldx = \iint_{D}(-\frac{\partial L}{\partial y})dA$ …..(1)

and

$\oint_{c}Mdy = \iint_{D}(\frac{\partial M}{\partial x})dA$ ….(2)

Here, the green’s theorem is proved in the first case.

The given diagram has the D region

D = {(x,y) | a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}

Here, g1 and g2 are continuous functions on [a, b].

Now, calculate the double integral in (1)

$\iint_{D}(\frac{\partial L}{\partial y})dA = \int_{a}^{b}\int_{g1(x)}^{g2(x)}\frac{\partial L}{\partial y}dydx$

=$\int_{a}^{b}L(x, g2(x))-L(x, g1(x))$

Now, calculate the line integral (I). From the diagram, C is written as C1, C2, C3, C4.

With C1,

$\int_{c_{1}}L(x,y)dx =\int_{a}^{b}L(x, g1(x))dx$ …….(3)

With C3,

$\int_{c_{3}}L(x,y)dx =-\int_{-c_{3}}L(x, y)dx$

= $=-\int_{a}^{b}L(x, g2(x))dx$

Therefore, C3 goes in the negative direction from b to a

Now, C2 and C4

Therefore,

$\int_{C}Ldx = \int_{c_{1}}L(x,y)dx + \int_{c_{2}}L(x,y)dx +\int_{c_{3}}L(x,y)dx + \int_{c_{4}}L(x,y)dx$

Therefore, the above expression is equal to

$\int_{a}^{b}L(x,g2(x))dx + int_{a}^{b}L(x,g1(x))dx$ …..(4)

Therefore, by combining (3) and (4), we get (1)

Hence, Proved.

## Green’s Theorem Area

With the help of Green’s theorem, it is possible to find the area of the closed curves.

From Green’s theorem,

$\oint_{C}(Ldx+Mdy)= \iint_{D}(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial x})dxdy$

If in the formula, $(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial x})$ = 1, then we have,

$\oint_{C}(Ldx+Mdy)= \iint_{D}dxdy$

Therefore, the line integral defined by the Green’s theorem gives the area of the closed curve. Therefore, we can write the area formulas as:

• $A = -\int_{c}ydx$
• $A = \int_{c}xdy$
• $A = \frac{1}{2}\int_{c}(xdy-ydx)$

## Green Gauss Theorem

If Σ is the surface Z which is equal to the function f(x, y) over the region R and the Σ lies in V, then

• $\iint_{\sum }P(x, y, z)d\sum$ exists.
• $\iint_{\sum }P(x, y, z)d\sum =\iint_{R}P(x, y, f(x,y))\sqrt{1+f_{1}^{2}(x,y)+f_{2}^{2}(x,y)}ds$

It reduces the surface integral to an ordinary double integral.

Green’s Gauss theorem can be stated from the above expression.

If P(x, y, z), Q(x, y, z), and R((x, y, z) are the three points on V, and it is bounded by the region $\sum^{\ast }$ and α, β, and γ are the direction angles, then

$\int \iint_{V}[P_{1}(x, y,z)+Q_{2}+R_{3}(x, y, z)]dV=\iint_{\sum^{\ast }}[P(x, y, z)cos\alpha + Q(x, y, z)cos\beta +R(x, y, z)cos\gamma ]d\sum$

### Green’s Theorem Example

Question:

Solve $\oint_{c}y^{3}dx-x^{3}dy$ where c is a circle of radius 2 centered in origin.

Solution:

Given: $\oint_{c}y^{3}dx-x^{3}dy$

Identify P and Q from the line integral

Here, P = y3 and Q= -x3

So,

$\oint_{c}y^{3}dx-x^{3}dy$ = $\iint_{D}-3x^{2}-3y^{2}dA$

= $-3\int_{0}^{2\pi }\int_{0}^{2}r^{3}drd\theta$

=$-3\int_{0}^{2\pi }\frac{1}{4}[r^{4}]_{0}^{2}d\theta$

=$-3\int_{0}^{2\pi }4d\theta$

= $-3[4]_{0}^{2\pi}$

= $-3(8\pi )$ = $-24\pi$

Therefore, $\oint_{c}y^{3}dx-x^{3}dy$ = $-24\pi$

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