Suppose that I have a double integral where the integrand has variables $x$ and $y$, and I am using the polar substitution $x=r\cos(\theta)$ and $y=r\sin(\theta)$. Suppose the region of integration is $x^2 + y^2 \leq 1$. By calculating the Jacobian $J = |r|$ I can replace the $dxdy$ by $|r|drd\theta$. I understand why the limits for $\theta$ will be $0$ to $2\pi$.
Here is where I am a bit confused. I know that $x^2 + y^2 = r^2 \leq 1 \implies -1 \leq r \leq 1$, and so this suggests $-1$ and $1$ will be the limits for $r$. But if I say that $r$ is always positive (thinking of it as the 'radius'), then it ranges from $0$ to $1$, and I can also say that the Jacobian $J = r$ instead of using the modulus. This is the approach I see people take when they do these kinds of integrals.
So, does the value of the integral change depending on whether I use $J = r$ and limits $0 \leq r \leq 1$, or use $J = |r|$ and limits $-1 \leq r \leq 1$? If it does change (and I am guessing that the non-modulus way is correct) then why is this so? Whilst I get that if I think about it intuitively, the 'radius' is always positive, I still don't see what is mathematically wrong about using $|r|$ and the $-1 \leq r \leq 1$ limits instead.
If you are using $r\in[-1,1]$, then each point $(x,y)$ in that disk can be written in two different ways as $(r\cos\theta,r\sin\theta)$. Therefore (using $\lvert r\rvert$) you will get twice the integral that you're after.