Try to solve that Question of Green theorm

**king imran** · June 11th 2007, 05:47 PM

"See Atach File"

**ThePerfectHacker** · June 11th 2007, 06:07 PM

$\oint_C (y^2dx + x^2 dy) = \iint_D \left( \frac{\partial x^2}{\partial x} - \frac{\partial y^2}{\partial y} \right) \ dA =\iint_D (2x - 2y) dA = 2\int_0^1 \int_0^{1-x} x - y \ dy \ dx$

You should be able to do it frum heir.

**topsquark** · June 11th 2007, 06:11 PM

[quote]Find $\oint_C y^2 \, dx + x^2 \, dy$ where the path is the boundary of the triangle given by x = 0, y = 0, and x + y = 1,[\quote]

First we need to define our area of integration. This is a right triangle with verticies at (0, 0), (1, 0), and (0, 1).

I will write
$\oint_C y^2 \, dx + x^2 \, dy = \oint_C \bold{F} \cdot d \bold{s}$
where $\bold{F} = \left [ \begin{matrix} y^2 \\ x^2 \end{matrix} \right ]$

Green's theorem says that
$\oint_C \bold{F} \cdot d \bold{s} = \int_A \nabla \times \bold{F} \cdot d \bold{a}$

So we first find $\nabla \times \bold{F}$ :
$\nabla \times \bold{F} = \left | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ y^2 & x^2 & 0 \end{matrix} \right |$ $= \hat{k} \left ( \frac{\partial (x^2)}{\partial x} - \frac{\partial (y^2)}{\partial y} \right ) = 2(x - y) \hat{k}$

Now, the area of this region is in the $- \hat{k}$ direction so:
$\oint_C y^2 \, dx + x^2 \, dy = -2 \int_A (x - y)da = -2 \int_0^1 \int_0^{-x + 1}(x - y) \, dx \, dy = 0$

-Dan

**ThePerfectHacker** · June 11th 2007, 06:58 PM

Originally Posted by topsquark

Green's theorem says that
$\oint_C \bold{F} \cdot d \bold{s} = \int_A \nabla \times \bold{F} \cdot d \bold{a}$

I see you learned Green's theorem. But there is a problem.

Actually Green's theorem says that:
$\iint_A (\nabla \times \bold{F})\cdot \bold{k} \ dA$ .

You forgot the normal outward pointing vector $\bold{k}$ .

(Unless the notation $d\bold{a}$ is so strange concept in Mathematical Physics. And it means $\bold{k}\cdot dA$ .)

**topsquark** · June 12th 2007, 04:24 AM

Originally Posted by ThePerfectHacker

(Unless the notation $d\bold{a}$ is so strange concept in Mathematical Physics. And it means $\bold{k}\cdot dA$ .)

Actually, yes!

-Dan

**ThePerfectHacker** · June 12th 2007, 11:01 AM

Originally Posted by topsquark

Actually, yes!

Why am I not supprised?

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