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I'm working on some odd problems in an older Larson calculus textbook to learn about vector fields. A (planar) vector field $\textbf F$ is called conservative if there exists a differentiable function $f$ such that $\textbf F(x,y)=\nabla f = \frac{\partial f}{\partial x} \textbf i + \frac{\partial f}{\partial y} \textbf j= M \textbf i+N \textbf j$.
One way to know if a vector field $\textbf F$ is conservative is iff $\frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}$, equivalent to saying that the mixed partial derivatives of the potential function $f$ are equal.

My specific problem says:

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The provided solution states:

enter image description here

I don't understand why the solution's first step says $M=\frac{\partial f}{\partial y}$ and $N=-\frac{\partial f}{\partial x}$. That doesn't fit with my introductory description of a vector field above. The rest of the solution flows from this initial step, but it seems like an initial contradictory leap out of the blue. The solution also doesn't use any mixed partials. Is this just a contrived function $f$ for this particular problem to make the vector field harmonic?

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    $\begingroup$ You've written down something that is quite wrong. If $\mathbf F$ is conservative, then $\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$. But the converse is false. This is not an "iff" statement. $\endgroup$ – Ted Shifrin Aug 20 '16 at 3:36
  • $\begingroup$ @TedShifrin: of course you're right, but I believe it is not the OP who wrote this, but the solution manual the OP is citing. I think this issue will be glossed over in "an older Larson calculus textbook," where it will always be assumed the region of integration is simply connected. $\endgroup$ – symplectomorphic Aug 20 '16 at 3:39
  • $\begingroup$ @symplectomorphic: That is terrible, terrible, terrible. I hope the OP knows better. $\endgroup$ – Ted Shifrin Aug 20 '16 at 3:41
  • $\begingroup$ @TedShifrin: whoops, I see the OP did write "iff" at the beginning of the post. My apologies; I was looking at the solution OP quoted. Agreed, though. $\endgroup$ – symplectomorphic Aug 20 '16 at 3:42
  • $\begingroup$ I'm assuming that the homework question, as posted, actually assumes $f$ is harmonic in the entire plane. Otherwise, of course, the statement is false. $\endgroup$ – Ted Shifrin Aug 20 '16 at 3:44
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You are missing the point. The line integral in the theorem you are trying to prove, namely

$$\int_cf_y\,dx-f_x\,dy\,,$$

is just another way of writing the line integral of $\mathbf{F}=\langle M,N\rangle$ over $c$, when $M=f_y$ and $N=-f_x$.

So the solution is saying assume $\mathbf{F}$ has that form. The trick is to then use Green's theorem (which is what Theorem 15.7 must be), which says

$$\int_c\mathbf{F}\cdot d\mathbf{r}=\iint_R\left(N_x-M_y\right)\,dA$$

where $R$ is the region enclosed by $c$.

But because $M=f_y$ and $N=-f_x$, we have $N_x=-f_{xx}$ and $M_y=f_{yy}$, so the right-hand side is

$$\iint_{R}(-f_{xx}-f_{yy})dA=\iint_{R}0dA=0$$

by the hypothesis that $f$ is harmonic.

ADDED: Actually, I did not read the proof in its entirety. I see it points out that $\mathbf{F}$ is conservative. This is true, so Theorem 15.7 might actually instead state the Fundamental Theorem of Line Integrals, which says that if $\mathbf{F}=\nabla f$, then

$$\int_c\mathbf{F}\cdot d\mathbf{r}=f(c(t_2))-f(c(t_1))$$

where $c(t_2)$ is the endpoint of $c$ and $c(t_1)$ is the initial point of $c$.

The result can also be proved this way, rather than with Green's theorem. The fact that $c$ is closed implies $c(t_2)=c(t_1)$, and the result follows.

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  • $\begingroup$ Note that either way -- via Green's theorem or via the Fundamental Theorem of Line Integrals -- you are using a version of Stokes's Theorem. $\endgroup$ – symplectomorphic Aug 20 '16 at 3:30
  • $\begingroup$ Oh -- I need to assume F has that form! I was trying to fit F into too rigid a box. (Theorem 15.7 just states some equivalent conditions: saying that F is conservative is equivalent to stating that the line integral of F.dr is independent of path, which is equivalent to stating that the integral of F.dr=0 for every closed curve C in R... This problem was one of the last problems in this section; Green's Theorem and Stoke's Theorem are in the next couple sections. Thank you for your explanation. $\endgroup$ – DBS Aug 20 '16 at 3:45
  • $\begingroup$ @DBS: you're welcome. Note that one part of that theorem (the part that says $F$ conservative implies its line integral around any closed curve is zero) uses the Fundamental Theorem of Line Integrals. $\endgroup$ – symplectomorphic Aug 20 '16 at 3:53
  • $\begingroup$ I see that now, thank you. That's Theorem 15.5. Slowly putting pieces together... $\endgroup$ – DBS Aug 20 '16 at 3:59
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The function $f:U\subset\Bbb{R}^2\to \Bbb{R}$ is given, all we know about it is that it's harmonic. $M=\frac{\partial f}{\partial y}$ is a function from a subset of $\Bbb{R}^2$ to $\Bbb{R}$, as is $N=-\frac{\partial f}{\partial x}$. Therefore, $M\, dx+ N\, dy$ is an ordinary vector field; you put in two numbers, you get back a vector with two components.

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