- #1

- 6

- 0

The question provides the vector field [itex](xy, 2yz, 3zx)[/itex] and asks me to confirm Stokes' theorem (the vector calc version) but I am trying to use the generalized differential forms version. So, I am trying to integrate [itex]\omega = xy\,dx + 2yz\,dy + 3zx\,dz[/itex] along the following triangular boundary [itex]\partial \Sigma[/itex]:

First, I try to find [itex]\int_\Sigma d\omega[/itex]. I computed [itex]d\omega = x\,dy\,dx + 2y\,dz\,dy + 3z\,dx\,dz[/itex] which simplifies to [itex]2y\,dz\,dy[/itex] on [itex]\Sigma[/itex]. However, integrating over the appropriate domain of [itex]y,z[/itex] with that differential gives 8/3 when the answer should be -8/3. I suspect my mistake is in not first reordering [itex]2y\,dz\,dy[/itex] to [itex]-2y\,dy\,dz[/itex] before integrating, but I'm not clear on the rational behind doing so. I am missing some intuition on what [itex]2y\,dz\,dy[/itex] and [itex]-2y\,dy\,dz[/itex] "are." In any case, where have I gone wrong?

First, I try to find [itex]\int_\Sigma d\omega[/itex]. I computed [itex]d\omega = x\,dy\,dx + 2y\,dz\,dy + 3z\,dx\,dz[/itex] which simplifies to [itex]2y\,dz\,dy[/itex] on [itex]\Sigma[/itex]. However, integrating over the appropriate domain of [itex]y,z[/itex] with that differential gives 8/3 when the answer should be -8/3. I suspect my mistake is in not first reordering [itex]2y\,dz\,dy[/itex] to [itex]-2y\,dy\,dz[/itex] before integrating, but I'm not clear on the rational behind doing so. I am missing some intuition on what [itex]2y\,dz\,dy[/itex] and [itex]-2y\,dy\,dz[/itex] "are." In any case, where have I gone wrong?