Why aren't $\int_0^\pi\int_{-1}^1e^rdr\,d\theta$ and $\int_0^{2\pi}\int_0^1e^rdr\,d\theta$ equal? Doesn't this violate the Change of Variables thm? Why aren't these two integrals equal?
$$\int_0^\pi \int_{-1}^{1} e^r \,dr\,d\theta \qquad\neq\qquad\int_0^{2\pi} \int_{0}^{1} e^r \,dr\,d\theta$$
Let me explain why I'm asking.
This is the change of variables theorem for double integrals:

Now, suppose that we have the unit disc $D \subset R^2$ and the transformation $T$ given by $x=r\cos\theta$ and $y=r\sin\theta$. Then the rectangle in the $r\theta$-plane $-1 \leq r \leq 1, 0 \leq \theta < \pi$ maps injectively to the unit disc under $T.$
So in theory, it seems like we should be able to integrate in polar coordinates using this region $-1 \leq r \leq 1, 0 \leq \theta < \pi$, in addition to the "usual" region $0 \leq r \leq 1, 0 \leq \theta < 2\pi$.

Then why aren't the above two integrals equal, and more importantly, why does this not violate the change of variables theorem?

 A: 
Why aren't these two integrals equal? $$\int_0^\pi \int_{-1}^{1} e^r \,\mathrm dr\,\mathrm d\theta \ne \int_0^{2\pi} \int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta\tag1$$

The common integrand $e^r$ does not vary the same way over the two different integration domains (let's call them $S_1$ and $S_2,$ respectively), which merely have the same measure and geometric representation. Consequently, the two integrals are not guaranteed to be equal. Indeed, taking their difference: \begin{align}&\int_0^{2\pi} \int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta
-\int_0^\pi \int_{-1}^{1} e^r \,\mathrm dr\,\mathrm d\theta\\
={}& \left(\int_0^\pi \int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta+\int_\pi^{2\pi} \int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta\right)
-\left(\int_0^\pi \int_{-1}^{0} e^r \,\mathrm dr\,\mathrm d\theta+\int_0^\pi \int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta\right)\\
={}&\int_\pi^{2\pi} \int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta-\int_0^\pi \int_{-1}^{0} e^r \,\mathrm dr\,\mathrm d\theta\\
={}& \pi\left(\int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta
-\int_{-1}^{0} e^r \,\mathrm dr\,\mathrm d\theta\right)\\
={}&\pi\,(1.72-0.63)\\
\ne{}&0.\\\end{align}

So in theory, it seems like we should be able to integrate in polar coordinates using this region $-1 \leq r \leq 1, 0 \leq \theta < \pi$

Only $S_2,$ but not $S_1,$ is in polar coordinates.
We can consider the entire inequation $(1)$ to be residing in a “coordinate system” that is simply not isomorphic to $\mathbb R^2$ (and thus maps the given geometric region to multiple integration domains). As such, notice that to fill in this blank $$\int_0^{2\pi} \int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta = \int_0^\pi \int_{-1}^{1} \fbox{$\phantom{filler}$}\,\mathrm dr\,\mathrm d\theta,$$ just replace each instance of $r$ with $|r|.$

why does this not violate the change of variables theorem?

This theorem isn't necessary here, but can be invoked via \begin{align}x&\color{red}=r\left|\cos\theta\right|,\\y&\color{red}=r\sin\theta,\\&f\Big(g(r,\theta),h(r,\theta)\Big)\det \left| \frac{\partial(x,y)}{\partial(r,\theta)} \right|\color{red}=e^{|r|}\ne e^r;\end{align} then $$\int_0^{2\pi} \int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta
\\=\int_0^{\frac\pi2} \int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta
+\int_{\frac\pi2}^{\frac{3\pi}2} \int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta
+\int_{\frac{3\pi}2}^{2\pi} \int_{0}^{1} e^r \,\mathrm dr\,\mathrm d\theta
\\\color{red}=\int_0^{\frac\pi2} \int_{0}^{1} e^{|r|} \,\mathrm dr\,\mathrm d\theta
+\left(\int_{\frac\pi2}^\pi \int_{0}^{1} e^{|r|} \,\mathrm dr\,\mathrm d\theta
+\int_{0}^{\frac{\pi}2} \int_{-1}^{0} e^{|r|} \,\mathrm dr\,\mathrm d\theta\right)
+\int_{\frac{\pi}2}^{\pi} \int_{-1}^{0} e^{|r|} \,\mathrm dr\,\mathrm d\theta
\\=\int_0^\pi \int_{-1}^{1} e^{|r|} \,\mathrm dr\,\mathrm d\theta.$$
A: Let's start from the "usual" region version and work backwards to get an integral in terms of $x$ and $y.$ One simple way to do this with the polar transformation we're used to is to simply factor out an $r$ for our Jacobian:
$$\begin{aligned}
\int_0^{2\pi} \int_0^1 e^r dr d\theta 
&= \iint_R \frac{e^r}{r} r dr d\theta \\
&= \iint_D \frac{e^{\sqrt{x^2 + y^2}}}{\sqrt{x^2 + y^2}} dy dx
\end{aligned}$$
noting that we can say $r = \sqrt{x^2+y^2}$ here because $r$ is always positive in our region.
However, if we want to solve this same integral using the other parametrization of the unit disc then we have to note that because $r$ is negative over some parts of the region, we have to use $\sqrt{x^2 + y^2} = |r|$ so we end up with a slightly different integrand:
$$\begin{aligned}
\iint_D \frac{e^{\sqrt{x^2 + y^2}}}{\sqrt{x^2 + y^2}} dy dx 
&= \iint_{R_2} \frac{e^{|r|}}{|r|} |r| dr d\theta\\
&= \int_0^{\pi} \int_{-1}^1 e^{|r|} dr d\theta
\end{aligned}$$
noting that the absolute value around the determinant of the Jacobian can also not be dropped in this case.
Now, noting that the integrand is even in $r$, we will see that the result of this integral will match the result of the first.
$$\begin{aligned}
\int_0^{\pi} \int_{-1}^1 e^{|r|} dr d\theta &= \left(\int_0^\pi d\theta\right)\left(\int_{-1}^1 e^{|r|} dr\right)\\
&= \pi \left(2 \int_0^1 e^{|r| }dr\right)\\
& = \left(\int_0^{2\pi} d\theta\right)\left(\int_0^1 e^r dr\right)\\
& = \int_0^{2\pi}\int_0^1 e^r dr d\theta
\end{aligned}$$
So ultimately, the reason that the two proposed integrals don't match is simply that they don't correspond to each other.
