Verify Green’s theorem in the plane for ∫c{(xy + y2)dx + x2dy} where C is the closed curve of the region bounded by y = x and y = x2.

$Now,\ we\ have\ to\ find\ the\ intersection\ of\ y=x\ and\ y=x^2.$

$RHS = x = x^2 $

$ x - x^2 = 0$

$ \Rightarrow x (1-x)= 0$

$i.e.\ x = 0,1$

$Hence,\ (0,0)\ and\ (1,1)\ will\ be\ the\ points\ of\ intersections.$

$According\ to\ Green's\ theorem,$

$\begin{aligned} \int_{c} M\ dx + N\ dy = \int_{} \int_{R} \left(\frac{\partial N}{\partial x} -\frac{\partial M}{\partial y}\right) d x \ d y \\ & \end{aligned}$

$According\ to\ the\ question,\ $

$\begin{aligned} \int_{c} (xy + y^2)\ dx + x^2\ dy = \end{aligned}$

$ = \begin{aligned} \int_{OA} ((xy + y^2)\ dx + x^2\ dy)\ + \int_{AO} ((xy + y^2)\ dx + x^2\ dy)\ = \end{aligned}$

$ = I_1 + I_2$

$Along\ OA,\ we\ have\ y\ = x^2,$

$\therefore dy\ = 2xdx\ and\ x\ varies\ from\ 0\ to\ 1$

$ I_1 = \begin{aligned} \int_{x=0}^{1} (x . x^2 + x^4)\ dx + x^2.2x\ dx \end{aligned}$

$ = \begin{aligned} \int_{x=0}^{1} (3x^3 + x^4) dx \end{aligned}$

$ = \begin{aligned} \left (\frac {3x^4} {4} + \frac {x^5}{5} \right)_0^{1} = \frac{3}{4} + \frac {1}{5} = \frac{19}{20}\end{aligned}$

$Along\ AO,\ we\ have\ y = x$

$\therefore\ dy = dx$

$x\ varies\ from\ 1\ to\ 0$

$ I_2 = \begin{aligned} \int_{1}^{0} (x . x + x^2)\ dx + x^2\ dx \end{aligned}$

$ = \begin{aligned} \int_{1}^{0} 3x^2 dx = [x^3]_1^{0} = -1 \end{aligned}$

$Hence,$

$L.H.S.\ =\ I_1 + I_2 = \frac{19}{20} - 1 = \frac{-1}{20}$

$Also,$

$R.H.S.\ =\ \begin{aligned} \int \int_{R} \left(\frac{\partial N}{\partial x} -\frac{\partial M}{\partial y}\right)\ dx\ dy \end{aligned}$

$where,\ N = x^2\ and\ M = xy + y^2$

$\Rightarrow \begin{aligned} \left (\frac{\partial N}{\partial x} \right) = 2x\ ,\ \left (\frac{\partial M}{\partial y} \right) = x + 2y \end{aligned}$

$R\ is\ the\ region\ bounded\ by\ y\ =\ x^2\ and\ y\ =\ x$

$\begin{aligned} \int_{} \int_{R} \left(\frac{\partial N}{\partial x} -\frac{\partial M}{\partial y}\right) dx\ dy \end{aligned}$

$ = \begin{aligned} \int_{x=0}^1 \int_{y=x^2}^{x} \left(2x - x- 2\right) dy\ dx \end{aligned}$

$ = \begin{aligned} \int_{x=0}^1 \int_{y=x^2}^{x} \left(x - 2y\right) dy\ dx \end{aligned}$

$ = \begin{aligned} \int_{x=0}^{1} \left(xy - y^2\right)_{y-x^2}^{x} dx \end{aligned}$

$ = \begin{aligned} \int_{x=0}^{1} \left((x^2 - x^2\right) - (x^3 - x^4)) dx \end{aligned}$

$ = \begin{aligned} \int_{x=0}^{1} \left(x^4 - x^3\right)dx \end{aligned}$

$ = \begin{aligned} \left(\frac {x^5}{5} - \frac {x^4}{4} \right)_0^{1} = (\frac {1}{5} - \frac {1}{4}) = \frac{-1}{20} \end{aligned}$

$\therefore L.H.S. = R.H.S. = \frac {-1}{20}$