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Please can someone verify my proof that $$\psi = {x dy - y d x\over x^2 + y^2}$$ is not exact?

Here is my work:

If $\psi$ was exact there would exist $f:\mathbb R^2 \setminus \{0\} \to \mathbb R$ such that $df = f_x dx + f_y dy = \psi$. Here $f_x = {-y \over x^2 + y^2}$ and $f_y = {x \over x^2 + y^2}$.

It would hold true that $$\int f_x dx = \int f_y dy = f.$$ So I calculate these integrals:

$$ \int f_x dx = -{1\over 2}\log(x^2 + y^2)$$

and

$$ \int f_y dy = {1\over 2}\log(x^2 + y^2)$$

It is clear that these cannot be equal therefore $\psi$ is not exact.

Edit

The first thing I had tried (it did not work) was to calculate the integral along a closed curve:

$$ \int_{S^1}\psi = \int_{S^1} {x \over x^2 + y^2} dy - \int_{S^1} {y \over x^2 + y^2} dx = \int_{S^1} x dy - \int_{S^1} y dx= x \int_{S^1} dy - y \int_{S^1} dx = 0$$

since $\int_{S^1}dx = 0$.

Arctic Char
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self-learner
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    If $\psi$ was exact on $\Bbb R^2 - {0} $, then the integral of $\psi $ along any closed loop in $\Bbb R^2 - {0} $ is zero. Can you find a loop $ c $ in $\Bbb R^2-{0} $ such that $\int_c \psi $ is nonzero? – kobe Dec 18 '14 at 02:38
  • @kobe Are you saying my solution is wrong? – self-learner Dec 18 '14 at 02:39
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    In your solution, $\int f_x, dx = \int f_y, dy = f$, but in fact $\int f_x, dx = f(x,y) + g(y)$ and $\int f_y, dy = f(x,y) + h(x)$ for some functions $g$ and $h$. The two integrals need not be the same. – kobe Dec 18 '14 at 02:47
  • Indeed. The thing to do here is apply the fact that a $1$-form $\omega$ is exact $\iff$ $\int_C \omega = 0$ for all closed curves $C$. – Kaj Hansen Dec 18 '14 at 02:52
  • I'm sorry but I still don't understand my mistake. Could someone please say a bit more? – self-learner Dec 18 '14 at 02:57
  • @kobe Why is the partial integral of $f_x$ with respect to $x$ $f(x,y) + g(y)$? Where does the function $g(y)$ come from? – self-learner Dec 18 '14 at 03:00
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    Your $f$ is a function of variables $x$ and $y$, so $\int f_x, dx$ differs from $f$ by a constant independent of $x$. This constant independent of $x$ must then be a function of $y$. So $\int f_x, dx = f(x,y) + g(y)$ for some function $g(y)$. – kobe Dec 18 '14 at 03:06
  • @kobe Why "must be" a function of $y$? Did you mean "can be"? – self-learner Dec 18 '14 at 03:15
  • @kobe Regarding your other comment: before I solved it like this I computed the integral along $S^1$ and it turned out to be zero. – self-learner Dec 18 '14 at 03:17
  • $$ \int_{S^1}\psi = \int_{S^1} {x \over x^2 + y^2} dy - \int_{S^1} {y \over x^2 + y^2} dx = \int_{S^1} x dy - \int_{S^1} y dx = 0$$ – self-learner Dec 18 '14 at 03:19
  • The partial integral $\int f_x, dx$ is a function of $x$ and $y$, so $g := \int f_x, dx - f$ is a function of $x$ and $y$ that is independent of $x$. Hence $g$ varies with $y$ alone. – kobe Dec 18 '14 at 03:29
  • In $\int_{S^1} x , \mathrm{d}y$, $x$ is not a constant, so you can't factor it out. –  Dec 18 '14 at 05:19
  • First things first. You need to learn the definition of a line integral. $\displaystyle\int_C x,dy = \int_a^b x(t)\dfrac{dy}{dt}dt$ when the curve is parametrized by $\big(x(t),y(t)\big)$, $a\le t\le b$. – Ted Shifrin Dec 18 '14 at 18:43
  • @TedShifrin Thank you, your comment helped. But isn't your formula missing an absolute value? I looked it up on Wikipedia and there the expression in the integral is $f(r) \cdot |r'|$. – self-learner Dec 18 '14 at 23:21
  • No, what you're talking about is a line integral of the form $\displaystyle\int_C f,ds$. – Ted Shifrin Dec 18 '14 at 23:51
  • @TedShifrin How did you go from $\int_C f ds$ to $\int_C x dy$? First, replace $f= f(x,y)=x$ in the integral and then? – self-learner Dec 19 '14 at 00:01

1 Answers1

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By Green's theorem,

$$\int_{S^1} \psi = \int_{S^1} x\, dy - y\, dx = \iint_{D^2} \text{div}(\langle x,y\rangle)\, dA = 2\cdot \text{Area}(D^2) = 2\pi$$

Alternatively, parametrize $S^1$ by setting $x = \cos(t)$, $y = \sin(t)$, $0 \le t \le 2\pi$. Then $$\int_{S^1} \psi = \int_0^{2\pi} (\cos(t)\cdot \cos(t) - \sin(t)\cdot (-\sin(t)))\, dt = \int_0^{2\pi} (\cos^2(t) + \sin^2(t))\, dt = 2\pi$$

Either way, $\int_{S^1} \psi \neq 0$.

kobe
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  • Thank you. But where is the mistake in $$ \int_{S^1}\psi = \int_{S^1} {x \over x^2 + y^2} dy - \int_{S^1} {y \over x^2 + y^2} dx = \int_{S^1} x dy - \int_{S^1} y dx = 0$$ I don't understand. – self-learner Dec 18 '14 at 03:43
  • You're correct up to the last step: $\int_{S^1} x, dy \neq \int_{S^1} y, dx$. – kobe Dec 18 '14 at 03:46
  • The integral $\int_{S^1} x, dy $ is positive, but $\int_{S^1} y, dx$ is negative. – kobe Dec 18 '14 at 03:47
  • I don't understand: $\int x dy = x \int dy$. And since $\int_{S^1} dx = 0$, $x \int dy = 0$ and $y \int dx = 0$. What is the mistake? – self-learner Dec 18 '14 at 04:08
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    You're misinterpreting the meanings of $\int_{S^1} x, dy$ and $\int_{S^1} y, dx$. These integrals are line integrals in the plane. The equation $\int_{S^1} x, dy = x\int_{S^1} dy$ does not make sense because $\int_{S^1} x, dy$ is constant and $x\int_{S^1} dy$ is variable. – kobe Dec 18 '14 at 04:26