VC5.7
From Exampleproblems
Here the parabola 
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 . 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\,](/wiki/images/math/0/b/e/0be9d22545f0a4ec79343771319c9f93.png)
RHS of (1) =
=![2\int_0^2 x[\frac{y^2}{2}+y]_{y=-\sqrt{8x}}^{\sqrt{8x}} dx\,](/wiki/images/math/d/8/8/d8843f6ee05d04f8ef44c43d1c520033.png)
on integrating w.r.t. y.
=![2\int_0^2 x[\frac{1}{2}(8x-8x)+\sqrt{8x}+\sqrt{8x}]dx\,](/wiki/images/math/b/8/5/b85fd051b4bfdbad701155f316dd18d8.png)
=![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}\,](/wiki/images/math/4/f/b/4fbec37d97bd54c89c91f7f7988cc1cc.png)
--(2)
Now,LHS=![\int_{AOB}[(x^2-2xy)dx+(x^2y+3)dy]+\int_{BA}[(x^2-2xy)dx+(x^2y+3)dy]\,](/wiki/images/math/4/0/3/4038a5dd83c7f128ac422def7ac5758e.png)
=![\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\,](/wiki/images/math/b/3/a/b3ab4e8765f05061fa05de89c8c9e095.png)
=![\int_{-4}^{4}[\frac{y^5}{256}-\frac{y^4}{16}+\frac{y^3}{64}+3]dy+\int_{-4}^{4}(4y+3)dy\,](/wiki/images/math/f/d/a/fdac281a50515d98e006c6b49703c8de.png)
=![-2\int_0^4 [3-\frac{y^4}{16}]dy+2(16-16)+3(4+4)\,](/wiki/images/math/4/2/e/42ea8be7e9fcc9ea6cb4d6b2449de0fb.png)
[Note that 
and 
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\,](/wiki/images/math/c/7/4/c7446f30e8e64ffd267c231d54bfc031.png)
=
on simplification. --(3)
From (2) and (3) LHS = RHS,hence Green's theorm is verified.
