# Math Insight

### When Green's theorem applies

Green's theorem provides another way to calculate \begin{align*} \dlint \end{align*} that you can use instead of calculating the line integral directly. However, some common mistakes involve using Green's theorem to attempt to calculate line integrals where it doesn't even apply.

First, Green's theorem works only for the case where $\dlc$ is a simple closed curve. If $\dlc$ is an open curve, please don't even think about using Green's theorem.

If you think of the idea of Green's theorem in terms of circulation, you won't make this mistake. Green's theorem converts the line integral to a double integral of the microscopic circulation. The double integral is taken over the region $\dlr$ inside the path. Only closed paths have a region $\dlr$ inside them. The idea of circulation makes sense only for closed paths. (And the line integral represents circulation around the region only if is simple.)

So if you are asked to compute the integral \begin{align*} \int_\dlc y\, dx + xy \, dy \end{align*} where $\dlc$ is the line from $(0,1)$ to $(1,1)$, can you use Green's theorem? No, because $\dlc$ is an open curve.

You could, for example, use Green's theorem to compute \begin{align*} \int_\dlc y\, dx + xy \, dy \end{align*} where $\dlc$ is the counterclockwise unit circle.

Second, Green's theorem can be used only for vector fields in two dimensions, such as the $\dlvf(x,y)= (y, xy)$ of the above example. It cannot be used for vector fields in three dimensions. So, don't bother with Green's theorem if you are given an integral like \begin{align*} \int_\dlc z\, dx + xy \, dy - yz \, dz \end{align*} even if $\dlc$ is a closed path.

Why a whole page just to stress these simple ideas? Because students frequently under pressure try to use Green's theorem when it doesn't apply. Maybe now that you've read this warning, you'll resist that temptation.