# VC5.7

Here the parabola $y^2=8x\,$ and the straight line x=2 intersect at the point say A(2,4) and B(2,-4). Thus the closed curve C consists of straight line BA and AOB of the parabola $y^2=8x\,$. Let R be the region bounded by C. Then,by Green's theorm in plane,we have

$\oint_C [(x^2-2xy)dx+(x^2y+3)dy]=\iint_R [\frac{\partial}{\partial x}(x^2y+3)-\frac{\partial}{\partial y}(x^2-2xy)]dx dy\,$

RHS of (1) =$\iint_R (2xy+2x)dx dy=2\int_{x=0}^{2}\int_{y=-\sqrt{8x}}^{\sqrt{8x}} x(y+1)dx dy\,$

=$2\int_0^2 x[\frac{y^2}{2}+y]_{y=-\sqrt{8x}}^{\sqrt{8x}} dx\,$ on integrating w.r.t. y.

=$2\int_0^2 x[\frac{1}{2}(8x-8x)+\sqrt{8x}+\sqrt{8x}]dx\,$

=$4\int_0^2 \sqrt{8}x^{\frac{3}{2}} dx=8\sqrt{2}[\frac{2x^{\frac{5}{2}}}{5}]_0^2=\frac{16}{5}\sqrt{2}2^{\frac{5}{2}}=\frac{128}{5}\,$ --(2)

Now,LHS=$\int_{AOB}[(x^2-2xy)dx+(x^2y+3)dy]+\int_{BA}[(x^2-2xy)dx+(x^2y+3)dy]\,$

=$\int_{y=4}^{-4}[(\frac{y^4}{64}-2\frac{y^2}{8}y)\frac{1}{4}ydy+(\frac{y^2}{64}y+3)dy]+\int_{-4}^{4}(4y+3)dy\,$

=$\int_{-4}^{4}[\frac{y^5}{256}-\frac{y^4}{16}+\frac{y^3}{64}+3]dy+\int_{-4}^{4}(4y+3)dy\,$

=$-2\int_0^4 [3-\frac{y^4}{16}]dy+2(16-16)+3(4+4)\,$ [Note that $\frac{y^5}{256}+\frac{y^3}{64}\,$ and $(3-\frac{y^4}{16})\,$ are respectively odd and even function of y]

=$-2[3y-\frac{1}{16}\frac{y^5}{5}]_0^4+24=-2[12-\frac{4^5}{80}]+24\,$

=$\frac{128}{5}\,$ on simplification. --(3)

From (2) and (3) LHS = RHS,hence Green's theorm is verified.

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