Question #22183

Verify Green's Theorem in the plane for integral_c [(x^2-xy^3)dx +(y^2-2xy)dy] where C is a square with vertices at (0,0), (2,0), (2,2), (0,2).
1

Expert's answer

2013-01-17T09:24:49-0500

Conditions

Verify Green's Theorem in the plane for integral_c [(x^2 - xy^3)dx + (y^2 - 2xy)dy] where C is a square with vertices at (0,0), (2,0), (2,2), (0,2).

Solution

As we know, the Green's Theorem claims:

Let CC be a positively oriented, piecewise smooth, simple closed curve in a plane, and let DD be the region bounded by CC. If LL and MM are functions of (x,y)(x, y) defined on an open region containing DD and have continuous partial derivatives there, then:


C(Ldx+Mdy)=D(MxLy)dxdy.\oint_{C} (L \, \mathrm{d}x + M \, \mathrm{d}y) = \iint_{D} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) \, \mathrm{d}x \, \mathrm{d}y.


Let's check the conditions of this theorem to check whether it works for our case.

As the C is a square with vertices at (0,0), (2,0), (2,2), (0,2), then this is positively oriented, piecewise smooth (because each side of this square could be represented as a linear function on a plane), simple closed curve. So, the conditions for C is completed.

Let's check the partial derivatives for functions:


L(x,y)=x2xy3L(x, y) = x^2 - x y^3M(x,y)=y22xyM(x, y) = y^2 - 2x y


As these functions are polynomials, then each their derivative is a continuous function. And these functions are defined in all R2R^2, so they are defined on each open region containing DD.

That's why the Green's Theorem is verified for our example.

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