Green’s theorem is a very important theorem of integration. We can also relate it with many theorem like strokes's theorem, gauss theorem. This theorem is mainly used for integration of a line combined with a curve plane. If we take line B and plane C, then we can find the combination of that integration*. *Green’s theorem is used in integrating derivatives in a particular plane. Here, we use basic theorem of integeration. By using green’s theorem, we will discuss some example problem.

Here, we will learn the proof of the green's theorem. Greeen's theorem shows the relationship between a line integral and a surface integral. If a line integral is given, we can convert it to surface integral or double integral and vise versa using this theorem.

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We can prove the Green's theorem as follows:

**Statement:**

$\oint_C (L dx + M dy) = \int \int_D (\frac{\partial M}{\partial x} - \frac{\partial L}{\partial x})dx dy$

**Proof:**

And,

$\int_C M dy = \int \int_d (\frac{\partial M}{\partial x}) dA$ ------------------(II)

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

The given diagram has the D region as,

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 (I)

$\int \int_D (\frac{\partial L}{\partial y}) dA = \int _a^b \int_{g_1(x)}^{g_2(x)} \frac{\partial L}{\partial y} dy dx$

= $\int_a^b L(x, g_2(x)) - L(x, g_1(x))$ -----------------(III)

Now, calculate the line integral (I). From the diagram, C is written as C_{1}, C_{2}, C_{3}, C_{4}.

With C_{1}

$\int_{C_1} L(x,y) dx = \int_a^b L (x, g_1(x)) dx$

With C_{3}

$\int_{C_3} L(x,y) dx = - \int_{-C_3} L(x,y) dx$

= $- \int_a^b L(x,g_2(x)) dx$

C_{3} goes in the negative direction from b to a.

Now, C_{2}, C_{4}

So,

$\int_C L dx = \int_{C_1} L (x,y) dx + \int_{C_2} L (x,y) dx + \int_{C_3} L (x,y) dx + \int_{C_4} L(x,y)$

= $\int_a^b L(x, g_2(x)) dx + \int_a^b L(x, g_1(x)) dx$ ----------------------(IV)

Combining (III) and (IV), we get (I). Similar computation gives (II).

With the help of Green's theorem, we can find the area of the closed curves. We know the formula for Green's theorem is$\oint _{C}(Ldx + M dy) = \int \int_{D}\left ( \frac{\partial M}{\partial x} - \frac{\partial N}{\partial y} \right )dx dy$

If in this formula, $\left ( \frac{\partial M}{\partial x} - \frac{\partial N}{\partial y}\right ) = 1$

Then, we get

$\oint _{C}(Ldx + M dy)= \int \int_{D}dx dy$

Then, the line integral defined by the Green's theorem gives the area of the enclosed region. So, the area

- A = $-\int _{C}y dx$
- A = $\int _{C}x dy$
- A = $\frac{1}{2}\int _{C}(x dy - y dx)$

Here, we use the formula $A = \frac{1}{2}\oint _{C}( xdy - ydx)$

where, C is the circle with the radius 2. Take x = 2 cos u, y = 2 sin u

where, 0 $\leq$ u $\leq$ 2$\pi$.

Then, the area is found as follows:

A = $\frac{1}{2}\left [ \int_{0}^{2\pi}(2 \cos u)(2 \cos u)du - \int_{0}^{2\pi}(2 \sin u)(2 \sin u)du \right ]$

= $\frac{1}{2}\int_{0}^{2\pi}4(\cos ^{2}u + \sin ^{2}u)du$

= $2\int_{0}^{2\pi}du$

= $4\pi$

where, C is the circle with the radius 2. Take x = 2 cos u, y = 2 sin u

where, 0 $\leq$ u $\leq$ 2$\pi$.

Then, the area is found as follows:

A = $\frac{1}{2}\left [ \int_{0}^{2\pi}(2 \cos u)(2 \cos u)du - \int_{0}^{2\pi}(2 \sin u)(2 \sin u)du \right ]$

= $\frac{1}{2}\int_{0}^{2\pi}4(\cos ^{2}u + \sin ^{2}u)du$

= $2\int_{0}^{2\pi}du$

= $4\pi$

If $\sum$ is the surface z = f(x,y) over the region R and $\sum$ lies in V, then

- $\int \int _{\sum}P(x,y,z)d\sum$ exsits
- $\int \int _{\sum}P(x,y,z)d\sum = \int \int _{R}P(x,y, f(x,y))\sqrt{1+f_{1}^{2}(x,y)+f_{2}^{2}(x,y)}dS$

This reduces a surface integral to an ordinary double integral.

With the use of above statement we can state the Green Gauss theorem as follows:

If P(x,y,z) and Q(x,y,z) and R(x,y,z) are the three points on V and V is bounded by the region $\sum ^{\ast }$ and $\alpha$ ,$\beta$, $\gamma$ are the direction angle of the outward normal to $\sum ^{\ast }$, then

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

Given below are some of the examples on Green's Theorem.

$\oint_c y^3 dx - x^3 dy$. Here, C is the positive oriented circle of radius 2 centered in origin.

Now, identify P and Q from line integral.

Here, $P = y^3$ and $Q = - x^3$.

$\oint y^3 dx - x^3 dy = \int \int_D -3x^2 - 3y^2 dA$

= $-3 \int_0^{2 \pi} \int_0^2 r^3 dr d \theta$

= $-3 \int_0^{2 \pi} \frac{1}{4}[r^4]_0^2 d \theta$

= $-3 \int_0^{2 \pi} 4 d \theta$

= $-3[4]_0^{2 \pi}$

= $-3(8 \pi)$

= $-24 \pi$

Evaluate $\oint_c$ x^{2}y^{2} dx + x^{2}y^{3} dy using Green's theorem where C is the triangle with vertices (0,0), (1,0), (1,2)

Draw C* *and D which makes a triangle.

The limits of x and y are as follows:

0 â‰¤ x â‰¤ 1, 0 â‰¤ y â‰¤ 2x

Here,

P = x^{2}y^{2} Q = x^{2}y^{3}

**Green's Theorem:**

$\oint_c$ x^{2}y^{2} dx + x^{2}y^{3} dy = $\int \int_D$ 2xy^{3 }- 2yx^{2} dA

= $\int_0^1 \int_0^{2x}$ 2xy^{3} - 2yx^{2} dy dx

= $\int_0^1$ 8x^{5 }- 4x^{4 }dx

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