Showing that $ \oint _C xdy-ydx=x_1y_2-x_2y_1$

Question

Let $C$ be the interval from point $(x_1,x_2)$ to point $(x_2,y_2)$

Show that $\displaystyle \oint _C xdy-ydx=x_1y_2-x_2y_1$

My attempt:

Acording Green's theoram $\displaystyle \oint _C xdy-ydx=\displaystyle \iint \bigg(\frac{\partial x}{\partial y}-\frac{-\partial y}{\partial x} \bigg)dxdy =\color{red}0$

Why I got zero? how can I solve this? any hints?

Answer 1

Hint: Use this parameterization of the line segment from point $(x_1,y_1)$ to point $(x_2,y_2)$:

$$x=x_1+(x_2-x_1)t$$ $$y=y_1+(y_2-y_1)t$$ $$0\le t\le 1$$

Then $dx=(x_2-x_1)dt,\ dy=(y_2-y_1)dt$, and calculating the integral is straightforward.

Answer 2

Hint: for the integral $$ \int_Cxdy-ydx, $$ write it as two separate integrals.

For $$ \int_Cxdy, $$ write $x$ as a function of $y$ (find the equation of the line and solve for $x$). Then use the definition of a line integral to evaluate.

The other integral is similar.

Answer 3

Thank's to @Rory Daulton I think I solved it:

$$\oint_C-ydx+xdy$$

$$\int\limits _{t=0}^{t=1}=-(y_1+t(y_2-y_1))(x_2-x_1)dt+\int\limits _{t=0}^{t=1}=(x_1+t(x_2-x_1))(y_2-y_1)dt$$

$$=-y_1(x_2-x_1)-\frac{(y_2-y_1)(x_2-x_1)}{2}+\frac{(x_2-x_1)(y_2-y_1)}{2}$$ $$=-y_1(x_2-x_1)+x_1(y_2-y_1)=\color{red}{\boxed{x_1y_2-x_2y_1}}$$