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What does it mean for an integral to be independent of a path? Take the path $x^4-6xy^3=4y^2$ for $$ \int_{(0, 0)}^{(2,1)} \left(10x^4-2xy^3\right)\,{dx}-3x^2y^2\, dy $$

My textbook says the integral is independent of the path since $$ \frac{\partial}{\partial y}\left(10x^4-2xy^3\right) = \frac{\partial}{\partial x}\left(3x^2y^2\right)\,{dy}. $$

But I've no idea what these partial derivatives (from Green's theorem) being equal has to do with it.

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3 Answers 3

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An integral is path independent if it only depends on the starting and finishing points.

Conservative fields $F$ are path independent, as they can be written as a gradient of a function $f$: $$ F=\nabla f $$ Consequently, on any curve $C=\{ r(t)\;|\; t\in [a,b]\}$, by the fundamental theorem of calculus $$ \int_C F\, dr = \int_C \nabla f \, dr = f(r(b))-f(r(a)), $$ in other words the integral only depends on $r(b)$ and $r(a)$: it is path independent.

Now, to link this with your question, when the partial derivatives from Green theorem are equal, the field is conservative (provided it is defined on a simply connected domain). Therefore, in your case, the path integral is path independent indeed.

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Integrating a Gradient

Since $$ \nabla\!\left(2x^5-x^2y^3\right)=\left(10x^4-2xy^3,-3x^2y^2\right)\tag{1} $$ we get $$ \begin{align} \int_{(x_0,y_0)}^{(x_1,y_1)}\nabla\!\left(2x^5-x^2y^3\right)\cdot(\mathrm{d}x,\mathrm{d}y) &=\int_{(x_0,y_0)}^{(x_1,y_1)}\mathrm{d}\!\left(2x^5-x^2y^3\right)\\ &=\left.2x^5-x^2y^3\vphantom{\int}\right|_{(x_0,y_0)}^{(x_1,y_1)}\tag{2} \end{align} $$


Exact Differential

This is also saying, using the chain rule, that $$ \begin{align} \mathrm{d}\!\left(2x^5-x^2y^3\right) &=\frac{\partial}{\partial x}\left(2x^5-x^2y^3\right)\mathrm{d}x+\frac{\partial}{\partial y}\left(2x^5-x^2y^3\right)\mathrm{d}y\\ &=\left(10x^4-2xy^3\right)\mathrm{d}x+\left(-3x^2y^2\right)\mathrm{d}y\tag{3} \end{align} $$ which is to say that $\left(10x^4-2xy^3\right)\mathrm{d}x+\left(-3x^2y^2\right)\mathrm{d}y$ is an exact differential.


Equality of Mixed Partials

We can check that $(3)$ is an exact differential using the Equality of Mixed Partials. That is, since $$ \frac{\partial^2}{\partial x\,\partial y}\left(2x^5-x^2y^3\right)=\frac{\partial^2}{\partial y\,\partial x}\left(2x^5-x^2y^3\right)\tag{4} $$ we have $$ \frac\partial{\partial y}\overbrace{\left(10x^4-2xy^3\right)}^{\frac\partial{\partial x}\left(2x^5-x^2y^3\right)}=\frac\partial{\partial x}\overbrace{\left(-3x^2y^2\right)}^{\frac\partial{\partial y}\left(2x^5-x^2y^3\right)}\tag{5} $$ Thus, the fact that $(3)$ is an exact differential implies $(5)$. However, to show that $(5)$ implies that $(3)$ is an exact differential, we need to use Green's Theorem.

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"Independent of the path" means that it does not matter which path you take, it will always end up taking the same amount of work to get from 'A' and 'B'.

This is a critical component of Green's Theorem, as it requires the field to be conservative (another way to describe that a path is independent).

You find out if a field is path independent by what the book did in the example, i.e.

If $\frac{d}{dy} = \frac{d}{dx}$ (i.e. $\frac{d}{dy} - \frac{d}{dx} = 0$) then your field is conservative.

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  • $\begingroup$ Could you please clarify the last bit when you get $0$? $\endgroup$
    – user129566
    Apr 9, 2016 at 3:11
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    $\begingroup$ This makes no sense. Green's Theorem applies only to closed paths enclosing a region. $\endgroup$ Apr 9, 2016 at 4:07
  • $\begingroup$ Agreed. The integral along the path may or may not be 0. A combination of one path from (0,0) to (2,1) and another path from (2,1) to (0,0) will form a closed loop if the second path is carefully chosen. $\endgroup$ Apr 9, 2016 at 5:29

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