# question about integral of e^(-x^2)

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

• $rdr=\frac{1}{2}d(r^2)$ Jan 25 '17 at 6:31
• thank you, positrón0802 ^^ Jan 25 '17 at 6:40
• Alternatively you can set $-r^2=s$ and proceed from there. Jan 25 '17 at 7:13

## 1 Answer

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

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

• it's embarrassing, but, for the latter part, why is that so? Jan 25 '17 at 6:33
• The chain rule... and please don't be embarrassed. Jan 25 '17 at 6:34
• @Dr.MV +1 for kindly explaining. Cheers! Jan 25 '17 at 6:38
• @qbert Thank you! Much appreciate the up vote and nice comment. -Mark Jan 25 '17 at 6:39
• 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. Jan 25 '17 at 6:39