Consider the two-dimensional vector field $\mathbf{F}=(x^2-y, x+y^2)$, with $\mathbf{x}=(x,y)\in \mathbb{R}^2$.

  • Compute the divergence and the curl of $\mathbf{F}$.

  • Compute the line integral $\int_C \mathbf{F}\cdot d\mathbf{x}$ if $C$ is the straight line between $(0,-1)$ and $(0,1)$.

  • What is the value of the integral if $C$ is changed to the semi-circle $x^2 +y^2 =1$ and $x\geq 0$?
  • I get $\text{div}(\mathbf{F})=2x+2y$, $\text{curl}(\textbf{F})=1-(-1)=2$.
  • The first component is unchanged, so the integral should just be $$\int_{-1}^1 (0)+y^2 dy=2/3$$
  • (No orientation is specified, so we assume the clockwise orientation). The integral over the whole semi-circle is $$\int_{\partial S}\mathbf{F}\cdot d\mathbf{x} = \int_S \text{curl}(\textbf{F}) = \pi.$$ So it should be $\pi - \frac{2}{3}$.

Does that look right?

  • $\begingroup$ Curl F is a vector field not a scalar. $\endgroup$ – user118653 Jan 2 '14 at 6:04
  • $\begingroup$ I thought the 2d curl was a scalar...i.e. $N_y-M_x$. $\endgroup$ – Eric Auld Jan 2 '14 at 6:07
  • 1
    $\begingroup$ Yes. If you wish, you can view the 2d situation as embedded in $\mathbb{R}^3$. The original vector field has zero third component, and the curl has the first two components zero, and the third is the "scalar curl". $\endgroup$ – Daniel Fischer Jan 2 '14 at 6:17

No orientation is specified, so we assume the clockwise orientation

You should assume counterclockwise orientation. The curve shall go from $(0,-1)$ to $(0,1)$, like the other one.

Anyway, you have a sign error. You have correctly computed the surface integral, and thus the integral over the boundary, which is

  • the semicircle $C$ in the counterclockwise direction, and
  • the straight line segment $L$ from $(0,1)$ to $(0,-1)$.

So we have

$$\pi = \int_{\partial S} \mathbf{F}\cdot d\mathbf{x} = \int_C \mathbf{F}\cdot d\mathbf{x} + \int_L \mathbf{F}\cdot d\mathbf{x},$$


$$\int_C \mathbf{F}\cdot d\mathbf{x} = \pi - \int_L \mathbf{F}\cdot d\mathbf{x} = \pi - \left(-\frac23\right) = \pi + \frac23,$$

since the first integral was over the line segment in the other direction. You can easily check that by parameterising the semicircle.

If you choose the clockwise orientation for the semicircle, the result would be $-\pi - \frac23$.

  • $\begingroup$ Thank you! That's a good reminder that for Green's theorem I have to make sure I'm using the positive orientation. $\endgroup$ – Eric Auld Jan 2 '14 at 21:21

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