0
$\begingroup$

I was looking for an answer for this question, I found it here on math stackexchage, but there's something in the answer I did not understand. I tried to add a comment too the answer, but I couldn't, because I dont have reputation enough. I cannot see where that tanget came from:

$ \int \int_{\mathbb{R}}f(x,y)dxdy = 2\pi \int_{0}^{\infty}tg(t)dt = \frac{2\pi}{B^2} $

link of the answer below: https://math.stackexchange.com/a/385427/755029

Could you help me, please?

$\endgroup$
1
  • 1
    $\begingroup$ Polar coordinates. $\endgroup$
    – Robert Israel
    Mar 2, 2020 at 20:16

1 Answer 1

Reset to default
1
$\begingroup$

There's no tangent there. The abbreviation for the tangent is $\tan$, not $\operatorname{tg}$; also, when a user has a reputation of several thousand, it’s rather likely that they can properly typeset math and wouldn’t italicize a function name.

The integrand is $t\cdot g(t)$. The factor $t$ is the Jacobian for the transformation from Cartesian to polar coordinates.

$\endgroup$
1
  • $\begingroup$ aaahhh, got it! Sorry. Here in Brazil some mathematicians and Physicists use this abreviation tgg for tangent. That was my confusion. Thank you very Much. $\endgroup$
    – Stenio W. R. da Silva
    Mar 2, 2020 at 21:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .