Solve line integral

Question

Evaluate $$\int_C (x-y) dx+(x+y)dy$$ $C$: circle with centre at origin having radius $2$.

Attempt: I know how to do line integral but my confusion here is how to choose parameter for $x$ and $y$

Answer 1

The integral can be split in two parts which can be very good to recognize: $$\int_C (x-y) \, dx + (x+y) \, dy = \int_C (x \, dx+ y \, dy) + (x \, dy - y \, dx) \\ = \int_C \frac12 \, d(x^2+y^2) + (x \, dy - y \, dx)$$

The first part contains an exact differential and therefore evaluates to $0$ for a closed curve: $$\int_C \frac12 \, d(x^2+y^2) = 0.$$

The second part, $\int_C x \, dy - y \, dx,$ is also good to remember and recognize as $\int_C \mathbf{r} \times d\mathbf{r}$ in 2 dimensions. It can easily be evaluated by some of the technics found in the other answers. Another closely related integral to remember and recognize is $$\int_C \frac{x \, dy - y \, dx}{x^2+y^2}$$ which evaluates to $2\pi$ if $C$ is a Jordan curve enclosing origin, and to $0$ if $C$ does not enclose origin.

Answer 2

In the comments, a hint is given to evaluate this integral directly. Here's a different way to do it using Green's Theorem which also gives an easy solution.

We have:

$$\int_C P \ dx \ + \ Q \ dy = \int_R\bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\bigg) \ dA = \int_R2 \ d A =8 \pi$$

Since the region is just a circle of radius $2$, and so has area $4 \pi$.

Answer 3

"Lowbrow method ":

Try parametrizing the circle by $x=2\cos t $ and $y=2\sin t $, where $t\in [0,2\pi] $...

Then $dx=-2\sin t dt$ and $dy=2\cos t dt $.

So, the integral becomes $\int_0^{2\pi}4((\cos t -\sin t)(-\sin t) +(\cos t +\sin t)(\cos t)) dt= 4\int_0^{2\pi}(\cos^2 t+\sin^2 t) dt=4\int_0^{2\pi}dt=4\cdot 2\pi=8\pi$...