Question

Use the method of this example to calculate F · dr, C where F(x, y) = 2xyi + (y2 − x2)j (x2 + y2)2 and C is any positively oriented simple closed curve that encloses the origin.

Answer 1

Answer:

Using Green's theorem we have:

int F.dr = int int d/dx ((y^2-x^2)/(x^2+y^2)^2) - d/dy (2xy/(x^2+y^2)^2) =

(2x^3 - 6xy^2) / (x^2+y^2)^3 - (2x^3 - 6xy^2) / (x^2+y^2)^3 = 0

Therefore:

int F.dr = 0

Step-by-step explanation:

Answer 2

Answer:

Around a close curve  $\oint_c \vec F \cdot dr=0$ .

Step-by-step explanation:

Given information:

$F(x, y) = \frac{2xyi + (y^2-x^2)j }{(x^2 + y^2)^2}$

If function is in form of,

$\vec F=Pi+Qj$ and C is any positively oriented simple closed curve that encloses the origin.

Then,by use of Green's theorem

$\int\int_{R}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA= \oint_c \vec F \cdot dr$

By partial differentiation

$\frac{\partial Q}{\partial x}=$$\frac{2x^3-6xy^2}{(x^2+y^2)^3}$

$\frac{\partial P }{\partial y}=\frac{2x^3-6xy^2}{(x^2+y^2)^3}$

On substitution in Green's theorem,

We get,

$\int\int_{R}(\frac{2x^3-6xy^2}{(x^2+y^2)^3}-\frac{2x^3-6xy^2}{(x^2+y^2)^3})dA= \oint_c \vec F \cdot dr$

$\oint_c \vec F \cdot dr=0$

Hence, around a close curve  $\oint_c \vec F \cdot dr=0$ .

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