Confusion over why Green's Theorem is usable in one situation with closed circle and not the other.

55 Views Asked by At

As I understand it, if a region contains the origin, Green's Theorem cannot be applied. However, one question in my issued lecture notes appears to contradict this and another one appears to follow this rule.

In the first question from the lecture notes, Green's Theorem can be used in this question (and is used in the model workings):

1 - Evaluate the line integral $\oint\,(e^{4x}-y^3)\,dx+(\cos 3y+x^3)\,dy$ where $C$ is the positively oriented circle $\,\,x^2+y^2=1$.

But for some reason, it is not applicable in this following question:

2 - Find the exact value of the line integral $\oint{x+y\over x^2+y^2}\,dx-{x-y\over x^2+y^2}\,dy\,\,$ where $C$ is the positively oriented circle $\,\,x^2+y^2=4\,\,$ (Hint: cannot apply Green’s Theorem, since $(0, 0)$ is in the region bounded by the circle $C$).

But if $x^2 + y^2 = 4$ contains the origin, doesn't $x^2 + y^2 = 1$ also contain the origin? What makes Green's Theorem applicable in the first example question but not the second?

1

There are 1 best solutions below

0
On BEST ANSWER

Green's Theorem doesn't depend on whether the enclosed region contains the origin; it depends on whether the functions have continuous partial derivatives inside the region (strictly speaking, on whether the functions have continuous partial derivatives on some open set containing the boundary and the region).

Your first pair of functions are well-behaved over the whole of $\Bbb R^2$, so Green's Theorem applies. But the functions in your second pair are not even defined at the origin, so Green's Theorem fails for any region containing the origin.