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Integration on the punctured plane

Mathematics Asked by MathNewbie on November 24, 2021

Let $M = mathbb{R}^2backslash{0}$,
$$ alpha = frac{xcdot dy-ycdot dx}{x^2+y^2}$$
and $mathcal{C} = {(f(e^{itheta})costheta,f(e^{itheta})sintheta);thetainmathbb{S}^1}$, where $f:mathbb{S}^1to (0,1)$ is smooth. Let $j:mathcal {C}to M$ be the inclusion. Calculate
$$int_mathcal{C}j^*alpha.$$
I’m not sure if I’m doing this right, but I got:
$$j^*alpha = frac{(xcirc j)d(ycirc j)-(ycirc j)d(xcirc j)}{(xcirc j)^2+(ycirc j)^2} =… = dtheta$$
So is the answer $int_0^{ 2pi}dtheta = 2pi?$ Or is it$int_0^{1}dtheta = 1$ since $f$ maps the circle on to $(0,1)$?

One Answer

You did not obtain answers so far, because your description of the problem is somehow overloaded. The underlying domain of integration is an "abstract copy of $S^1$". This copy can be realized as ${mathbb R}/(2pi{mathbb Z})$ or as geometric unit circle $partial D$ in ${mathbb R}^2$. At any rate the chosen parameter variable is $theta$, and for a full performance through $S^1$ this $theta$ should run over the interval $[0,2pi]$. On the other hand it is true that this $theta$ is not a 100% coordinate function on $S^1$, but we have control over the details $ldots$

Understanding all this, your $f$, defined on the points of $partial D$, should be considered as a function $$rho(theta):=f(e^{itheta})in ]0,1[qquad(0leqthetaleq2pi) .$$ At any rate, your $j^*alpha=dtheta$ is correct; and $$int_{cal C} j^*alpha=int_0^{2pi}dtheta=2pi .$$ There is no question of an integral $int_0^1dtheta$ here.

Maybe it is of some help to note that $alpha=darg$, since $$nablaarg(x,y)quadleft(=nablaarctan{yover x}right)quad=left({-yover x^2+y^2}, >{xover x^2+y^2}right)$$ is valid for all $(x,y)ne(0,0)$.

Answered by Christian Blatter on November 24, 2021

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