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So here $M(x,y)=8x+3y$ and

$N(x,y)=g(x,y)+18y^2$

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=3$

Integrating $\frac{\partial N}{\partial x}$ with respect to $x$ and holding $y$ fixed

$N(x,y)=\int\frac{\partial N}{\partial x}=3x+h(y)$

differentiating $N(x,y)$ with respect to $y$

$\frac{\partial N}{\partial y}=\frac{\partial g}{\partial y}+36y$

I've noticed that however much i try integrating or differentiating either $M(x,y)$ or $N(x,y)$ and their partial derivatives, i always end up with two unknowns. Can someone help?

pkjag
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As all integrals of $\mathbf{F}$ around loops are $0$ on (the open, simply-connected) plane, you are right that $\mathbf{F}$ is conservative. (I'm assuming more than mere continuity here.) You are right in asserting $M_y = N_x = 3$ then follows. Since $\frac{\partial}{\partial x} 18y^2=0$ then $g(x,y)$ may take the form $g(x,y) = 3x + h(y)$ for any function $h$ of the single variable $y$.

Did anyone assert that the solution for the function $g$ had to be unique?

Randall
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$\require{begingroup}\begingroup \newcommand{\degs}{^\circ} \newcommand{\cut}{\setminus} \newcommand{\AND}{\ \ {\rm{\small{AND}}}\ \ } \newcommand{\OR}{\ \ {\rm{\small{OR}}}\ \ } \newcommand{\NOT}{\ \ {\rm{\small{NOT}}}\ \ } \newcommand{\Implies}{\Rightarrow} \newcommand{\If}{\Leftarrow} \newcommand{\Iff}{\Leftrightarrow} \newcommand{\x}{\times} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\E}{\operatorname{\rm{\small{E}}}} \renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}} \newcommand{\dash}{\textrm{-}} \newcommand{\der}{\partial} \newcommand{\del}{\nabla} \newcommand{\inv}{{\sim}} \newcommand{\eps}{\varepsilon} \newcommand{\indent}{\ \ \ \ \ \ } \newcommand{\dedent}{\!\!\!\!\!\!\!\!\!}$ I think it's actually okay that there are some unknowns.

The reason is there are many possible $g(x,y)$ functions that will solve this problem, i.e. that will satisfy the condition that the resulting vector field is conservative.

You correctly stated that $$ \frac{\der M}{\der y} = \frac{\der N}{\der x} $$ which simplifies to $$ 3 = g_x $$ Integrating both sides with respect to $x$ yields $$ g(x,y) = 3x + h(y) $$ where $h(y)$ is an arbitrary function.

At this point you might be tempted to try to somehow solve for $h(y)$ (I was, anyway), but it turns out this $h(y)$ is not something to solve for, it rather describes the form of all the possible $g(x,y)$ solutions: i.e. that as long as $g(x,y)$ has the form $$ g(x,y) = 3x + h(y) $$ for some function $h(y)$, then $g$ is a valid solution to the problem. If you don't believe me, try plugging in the above expression for $g(x,y)$ and solving for the potential function $f$ of the vector field $\vec{\mathbf{F}}$ in terms of $h(y)$. You'll see that it doesn't matter what $h(y)$ is, we can always find $f$.

$\endgroup$

WB-man
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  • OK so what if I've been asked to evaluate the integral $\int F(x,y)dr$ around an oriented curve starting from (1,4) to (0,3)? – pkjag Aug 11 '17 at 15:38
  • @pkjag I assume you're referring to a vector line integral $\int \vec F(x,y) \cdot d\vec r$. If $\vec F$ is a conservative vector field with potential function $f$, you can use the Fundamental Theorem of Line Integrals to evaluate it as $f(0,3)-f(1,4)$. If $\vec F$ is not conservative, you'll probably have to evaluate it directly. – WB-man Aug 12 '17 at 03:24
  • So I ignored $h(y)$ when finding the potential function $f$, is that accurate? – pkjag Aug 12 '17 at 11:35
  • @pkjag I don't think you can ignore $h(y)$. Since we're given no additional information about $\vec F$, we have to use the general formula for $g(x,y)$ which includes the arbitrary $h(y)$ function. When finding the potential function $f$, you'll need the antiderivative of $h(y)$ which I'll call $H(y)$. The value of the line integral will have to be in terms of $H(y)$. But note that, if instead of $(0,3)$ and $(1,4)$, the curve's endpoints had had the same $y$-coordinate, the $H$ functions would cancel, and then you could ignore $h(y)$. – WB-man Aug 12 '17 at 15:52
  • Then how should I evaluate the line integral? – pkjag Aug 12 '17 at 23:55
  • @pkjag I don't think you can evaluate the line integral into a single specific number. The value of the line integral will have to be expressed in terms of $H(3)$ and $H(4)$ unless we are given more information about the vector field $\vec F$. – WB-man Aug 13 '17 at 00:00