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. \[ \oint_C Mdx + Ndy = \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)dxdy \] 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 \[\hspace{5pt} \oint_C (xy - y^2)dx + x^2dy \], 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: \[ x = x^2 \] \[ x(x - 1) = 0 \] \[ \Rightarrow x = 0, x = 1 \] \[ \text{For } x = 0, \hspace{5pt} y = 0 \] \[ \text{For } x = 1, \hspace{5pt} y = 1 \] So the points of intersection are (0,0) and (1,1)
By Green's Theorem, \[ \oint_C Mdx + Ndy = \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)dxdy \] L.H.S. : \[ \oint_C (xy - y^2)dx + x^2dy \] \[ \int_{C_1} \left[(xy - y^2)dx + x^2dy \right] + \int_{C_2} \left[(xy - y^2)dx + x^2dy \right] \] Now for curve C1 : \[ y = x \] \[ dy = dx \] Points are (0,0) to (1,1) \[ = \int_{C_2} (xy - y^2)dx + x^2dy \] \[ = \int_1^0 (x^2 - x^2)dx + x^2dx \] \[ = \int^0_1 x^2dx \] \[ \left[\frac{x^3}{3} \right]^0_1 \] \[ = \boxed{\frac{-1}{3}} \] For curve C2, the equation is: \[ y = x^2 \] \[ \Rightarrow dy = 2xdx \] The points are: (0,0) to (1,1) \[ \int_C (xy - y^2)dx + x^2dy \] \[ = \int_0^1 \left[x(x^2) - (x^2)^2 \right]dx + x^2(2xdx) \] \[ = \int_0^1 \left(x^3 - x^4 + 2x^3 \right)dx \] \[ = \int_0^1 \left(3x^3 - x^4 \right)dx \] \[ = \left[\frac{3x^4}{4} - \frac{x^5}{5} \right]^1_0 \] \[ = \frac{3}{4} - \frac{1}{5} = \boxed{\frac{11}{20}} \] So, \[ = \oint (xy - y^2)dx + x^2dy \] \[ = \frac{11}{20} - \frac{1}{3} \] \[ = \boxed{\frac{13}{60}} \] R.H.S. : \[ \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)dxdy \] \[ = \iint_R \left[\frac{\partial}{\partial x}(x^2) - \frac{\partial}{\partial y}(xy - y^2) \right]dxdy \] \[ = \iint_R \left[2x - (x - 2y) \right]dxdy \] \[ = \iint_R (x + 2y)dxdy \] Now using double integral, here we will use horizontal strip to perform integration:
\[ \iint_R (x + 2y)dxdy \] \[ = \int_0^1 \int^{\sqrt{y}}_y (x + 2y)dxdy \] \[ = \int^1_0 \left[\frac{x^2}{2} + 2xy \right]^{\sqrt{y}}_{y} dy \] \[ = \int^1_0 \left[\frac{y}{2} + 2y^{3/2} - \frac{y^2}{2} - 2y^2 \right]dy \] \[ = \left[\frac{y^2}{4} + \frac{4y^{5/2}}{5} - \frac{5y^3}{6} \right]^1_0 \] \[ = \frac{1}{4} + \frac{4}{5} - \frac{5}{6} \] \[ = \boxed{\frac{13}{60}} \] L.H.S. = R.H.S., Hence Proved :)
Sample Problem 2
Using Green's Theorem \[ \oint_C (y - \sin{x})dx + \cos{x}dy \] 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 \[ \oint_C Mdx + Ndy = \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)dxdy \] \[ = \int^{\large\frac{\pi}{2}}_0 \int^{\large\frac{2}{\pi}x}_0 \left[\frac{\partial}{\partial x}(\cos{x}) - \frac{\partial}{\partial y}(y\sin{x}) \right]dxdy \] \[ = \int^{\large\frac{\pi}{2}}_0 \int^{\large\frac{2}{\pi}x}_0 \left[-\sin{x} - 1 \right]dxdy \] \[ = \int^{\large\frac{\pi}{2}}_0 \left[-y\sin{y} - y \right]^{\large\frac{2}{\pi}x}_0 dx \] \[ = \int^{\large\frac{\pi}{2}}_0 \left(\frac{-2x\sin{x}}{\pi} - \frac{2x}{\pi} \right)dx \] \[ = \frac{-2}{\pi} \int^{\large\frac{\pi}{2}}_0 (x\sin{x} + x)dx \] \[ = \frac{-2}{\pi} \left[x\frac{d}{dx}(\sin{x}) - \int \left(\frac{d(x)}{dx}\int \sin{x}dx \right)dx + \frac{x^2}{2} \right]_0^{\frac{\pi}{2}} \] \[ = \frac{-2}{\pi} \left[x\cos{x} + \sin{x} + \frac{x^2}{2} \right]_0^{\frac{\pi}{2}} \] \[ = \frac{-2}{\pi} \left[0 + 1 + \frac{\pi^2}{8} \right] + \frac{2}{\pi} \left[0 + 0 + 0 \right] \] \[ = \boxed{\frac{-8 + \pi^2}{4\pi}} \]
