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[quote]Find $\displaystyle \oint_C y^2 \, dx + x^2 \, dy$ 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
$\displaystyle \oint_C y^2 \, dx + x^2 \, dy = \oint_C \bold{F} \cdot d \bold{s}$
where $\displaystyle \bold{F} = \left [ \begin{matrix} y^2 \\ x^2 \end{matrix} \right ] $
Green's theorem says that
$\displaystyle \oint_C \bold{F} \cdot d \bold{s} = \int_A \nabla \times \bold{F} \cdot d \bold{a}$
So we first find $\displaystyle \nabla \times \bold{F}$:
$\displaystyle \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 | $ $\displaystyle = \hat{k} \left ( \frac{\partial (x^2)}{\partial x} - \frac{\partial (y^2)}{\partial y} \right ) = 2(x - y) \hat{k}$
Now, the area of this region is in the $\displaystyle - \hat{k}$ direction so:
$\displaystyle \oint_C y^2 \, dx + x^2 \, dy = -2 \int_A (x - y)da = -2 \int_0^1 \int_0^{-x + 1}(x - y) \, dx \, dy = 0$
-Dan
I see you learned Green's theorem. But there is a problem.
Actually Green's theorem says that:
$\displaystyle \iint_A (\nabla \times \bold{F})\cdot \bold{k} \ dA$.
You forgot the normal outward pointing vector $\displaystyle \bold{k}$.
(Unless the notation $\displaystyle d\bold{a}$ is so strange concept in Mathematical Physics. And it means $\displaystyle \bold{k}\cdot dA$.)