0
$\begingroup$

[enter image description here]

why does dr becomes dr^2? hope someone helps me~!

  • 2
    $rdr=\frac{1}{2}d(r^2)$ – 2017-01-25
  • 0
    thank you, positrón0802 ^^ – 2017-01-25
  • 0
    Alternatively you can set $-r^2=s$ and proceed from there. – 2017-01-25

1 Answers 1

3

Note that from the Jacobian matrix, the surface differential $dx\,dy$ transforms to $r\,dr\,d\theta$ in cylindrical coordinates.

Then note from the chain rule that $\frac12 d(r^2)=r\,dr$.

Putting it together yields

$$\int_{-\infty}^\infty \int_{-\infty}^\infty (\cdot)\,dx\,dy = \int_0^{2\pi}\int_0^\infty (\cdot)\,r\,dr\,d\theta=\int_0^{2\pi}\int_0^\infty (\cdot)\frac12 d(r^2)\,d\theta$$

  • 0
    it's embarrassing, but, for the latter part, why is that so? – 2017-01-25
  • 2
    The chain rule... and please don't be embarrassed. – 2017-01-25
  • 0
    @Dr.MV +1 for kindly explaining. Cheers! – 2017-01-25
  • 0
    @qbert Thank you! Much appreciate the up vote and nice comment. -Mark – 2017-01-25
  • 0
    now it's so obvious, I just couldn't think about using it solving this kind of integral. thank you! but why do you say it's about product rule? i see it d(r^2)/dr=2r. – 2017-01-25
  • 0
    Not the product rule; the chain rule. So $d(r^2)=2r\,dr$ and hence $\frac12d(r^2)=r\,dr$. – 2017-01-25
  • 0
    You're welcome. My pleasure. -Mark – 2017-01-25