Green's Theorem Engineering Maths, Btech first year

Green's Theorem

This theorem gives the relationship between Line Integral across a simple closed curve C and Double Integral over a plane region R bounded by that curve C. $${\oint }_{C}Mdx+Ndy={\iint }_R(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dxdy$$ Here $M$ and $N$ are functions of $x$ and $y$ and they have continuous partial derivatives.

Advantages of Green's Theorem

At L.H.S. is Line Integral which depends upon number of lines in closed curves and thus it is a long calculation. On the other hand, R.H.S. has Double Integral which depends number of regions. Generally, this number is 1, so the calculation becomes short. For example: for a triangle, its Line Integral will consist of 3 Line Integrals as there are three lines in a triangle, whereas its Double Integral will be done only one time as the number of regions is just 1.

Sample Problem 1

Verify Green's Theorem for $${\oint }_C(xy-y^2)dx+x^{2}dy$$, where C is bounded by curves $y=x$ and $y=x^2$.

Solution

The given curves are: $$\text{C\_1 }:y=x$$ $$\text{C\_2 }:y=x^2$$ Point of intersection: $$x=x^2$$ $$x(x-1)=0$$ $$\Rightarrow x=0,x=1$$ $$\text{For }x=0,y=0$$ $$\text{For }x=1,y=1$$ So the points of intersection are (0,0) and (1,1)

By Green's Theorem, $${\oint }_{C}Mdx+Ndy={\iint }_R(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dxdy$$ L.H.S. : $${\oint }_C(xy-y^2)dx+x^{2}dy$$ $${\int }_{C_1}[(xy-y^2)dx+x^{2}dy]+{\int }_{C_2}[(xy-y^2)dx+x^{2}dy]$$ Now for curve C₁ : $$y=x$$ $$dy=dx$$ Points are (0,0) to (1,1) $$={\int }_{C_2}(xy-y^2)dx+x^{2}dy$$ $$={\int }_1^0(x^2-x^2)dx+x^{2}dx$$ $$={\int }_1^0{x}^{2}dx$$ $${\left[\frac{x^3}{3}\right]}_1^0$$ $$=\boxed{{\displaystyle \frac{-1}{3}}}$$ For curve C₂, the equation is: $$y=x^2$$ $$\Rightarrow dy=2xdx$$ The points are: (0,0) to (1,1) $${\int }_C(xy-y^2)dx+x^{2}dy$$ $$={\int }_0^1[x(x^2)-(x^2{)}^2]dx+x^2(2xdx)$$ $$={\int }_0^1(x^3-x^4+2x^3)dx$$ $$={\int }_0^1(3x^3-x^4)dx$$ $$={[\frac{3x^4}{4}-\frac{x^5}{5}]}_0^1$$ $$=\frac{3}{4}-\frac{1}{5}=\boxed{{\displaystyle \frac{11}{20}}}$$ So, $$=\oint (xy-y^2)dx+x^{2}dy$$ $$=\frac{11}{20}-\frac{1}{3}$$ $$=\boxed{{\displaystyle \frac{13}{60}}}$$ R.H.S. : $${\iint }_R(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dxdy$$ $$={\iint }_R[\frac{\partial }{\partial x}(x^2)-\frac{\partial }{\partial y}(xy-y^2)]dxdy$$ $$={\iint }_R[2x-(x-2y)]dxdy$$ $$={\iint }_R(x+2y)dxdy$$ Now using double integral, here we will use horizontal strip to perform integration:

$${\iint }_R(x+2y)dxdy$$ $$={\int }_0^1{\int }_y^{\sqrt{y}}(x+2y)dxdy$$ $$={\int }_0^1{[\frac{x^2}{2}+2xy]}_y^{\sqrt{y}}dy$$ $$={\int }_0^1[\frac{y}{2}+2y^{3/2}-\frac{y^2}{2}-2y^2]dy$$ $$={[\frac{y^2}{4}+\frac{4y^{5/2}}{5}-\frac{5y^3}{6}]}_0^1$$ $$=\frac{1}{4}+\frac{4}{5}-\frac{5}{6}$$ $$=\boxed{{\displaystyle \frac{13}{60}}}$$ L.H.S. = R.H.S., Hence Proved :)

Sample Problem 2

Using Green's Theorem $${\oint }_C(y-\sin x)dx+\cos xdy$$ where C is the triangle enclosed by lines $$y=0,x=\frac{\pi }{2},y=\frac{2}{\pi }x$$

Solution

Tracing the curves by using given equations: $$y=0,x=\frac{\pi }{2},y=\frac{2}{\pi }x$$

Points of intersection : $$y=\frac{2}{\pi }x$$ $$x=\frac{\pi }{2}\Rightarrow y=\frac{2}{\pi }\times \frac{\pi }{2}$$ $$y=1$$ So the points of intersection are (0,0), ($\frac{\pi }{2}$,0) and ($\frac{\pi }{2}$,1) By Green's Theorem $${\oint }_{C}Mdx+Ndy={\iint }_R(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dxdy$$ $$={\int }_0^{\frac{\pi }{2}}{\int }_0^{\frac{2}{\pi }x}[\frac{\partial }{\partial x}(\cos x)-\frac{\partial }{\partial y}(y\sin x)]dxdy$$ $$={\int }_0^{\frac{\pi }{2}}{\int }_0^{\frac{2}{\pi }x}[-\sin x-1]dxdy$$ $$={\int }_0^{\frac{\pi }{2}}{[-y\sin y-y]}_0^{\frac{2}{\pi }x}dx$$ $$={\int }_0^{\frac{\pi }{2}}(\frac{-2x\sin x}{\pi }-\frac{2x}{\pi })dx$$ $$=\frac{-2}{\pi }{\int }_0^{\frac{\pi }{2}}(x\sin x+x)dx$$ $$=\frac{-2}{\pi }{[x\frac{d}{dx}(\sin x)-\int (\frac{d(x)}{dx}\int \sin xdx)dx+\frac{x^2}{2}]}_0^{\frac{\pi }{2}}$$ $$=\frac{-2}{\pi }{[x\cos x+\sin x+\frac{x^2}{2}]}_0^{\frac{\pi }{2}}$$ $$=\frac{-2}{\pi }[0+1+\frac{{\pi }^2}{8}]+\frac{2}{\pi }[0+0+0]$$ $$=\boxed{{\displaystyle \frac{-8+{\pi }^2}{4\pi }}}$$