I apologize in advance if this is a really stupid question.
So, I want to find the area of the region bounded by $\sqrt{1+x^2}$ and $-\sqrt{1+x^2}$ and $x=\pm 1$. I came up with the double integral: $$\int^1_{-1} \int^{\sqrt{1+x^2}}_{-\sqrt{1+x^2}} 1 \, dy\, dx $$ I then did the "intuitive transformation" of $x=iu$ and $y=v$. The Jacobian of this would be $|i\cdot 1 - 0\cdot 0|=|i|$, and so the integral would transform to $$\int^1_{-1} \int^{\sqrt{1-u^2}}_{-\sqrt{1-u^2}} |i| \, dv\, du$$ which would "intuitively" result int $\pi |i|$.
Well, the integral of that region is actually $2(\sqrt 2 + \operatorname{arcsinh}(1))\approx 4.6$, which is not anywhere related to $\pi$ or $i$.
Is this a really stupid misuse of the Jacobian? My professor did off-handedly mention that complex Jacobians were the same as real ones, but I'm not sure if he fully meant that or if I misunderstood. Is there a computation that can transform to the complex plane and get the correct answer?
Thanks for your time.
You've substituted $x=ui$,
At $x=-1, u = i$
$x=1, u = -i$
So, the integral would become,
$$I = \int^{-i}_i\int^{\sqrt{1-u^2}}_{\sqrt{1-u^2}}idvdu = 4i\int^{-i}_0\sqrt{1-u^2}du = 4i\bigg[\frac{u}{2}\sqrt{1-u^2}+\frac{1}{2}\sin^{-1}u\bigg]^{-i}_0 = 4i\bigg[\frac{i}{2}\sqrt2+\frac{1}{2}\sin^{-1}{i}\bigg] = 4i\bigg[\frac{-i\sqrt2}{2}+\frac{i}{2}\sinh^{-1}(-1)\bigg] = 2(\sqrt2+\sinh^{-1}1)$$