Double integral substitution where the region of integration becomes a point I want to evaluate a double integral of some function over the region between the graphs $y=x^2$ and $y=\frac{x^2}{2}+1$.
$D=\{(x,y):-\sqrt2\leq x \leq \sqrt2, x^2\leq y \leq \frac{x^2}{2}+1\}$
The subsitution I am trying to make is
$$x^2=u$$
$$y=v$$
The way I visualise this substitution is folding around the $y$ axis from left to right and making the curves straight lines.
$$y=x^2 \rightarrow v=u$$
$$y=\frac{x^2}{2}+1 \rightarrow v=\frac{u}{2}+1$$
From there I get confused. The Jacobian is always zero $\pm \frac{1}{2\sqrt{u}}$ and $D$ became a single point $(2,2)$. Doesn't this make the integral zero?
 A: $ \displaystyle D=\{(x,y):-\sqrt2\leq x \leq \sqrt2, x^2\leq y \leq \frac{x^2}{2}+1\}$
The integral is straightforward without change of variable. Coming to the change of variable that you are using,
$u = x^2, v = y$
The substitution you are using does not give one to one mapping between the original region and the region after transformation. In other words, note that substitution $u = x^2$ (given the bounds of $y$ is a function of $x^2$) leads to the original region for $x \lt 0$ and $x \gt 0$ map to the same region after change of variable. So you need to be careful while integrating a function over the given region. See the second example below.
The bounds are $0 \leq u \leq 2, u \leq v \leq \frac u 2 + 1$
Jacobian $|J| = \frac{1}{2 \sqrt u}$
So for example, if we were finding bound area of the region,
$ \displaystyle \int_{- \sqrt 2}^{\sqrt 2} \int_{x^2}^{\frac {x^2} 2 + 1} 1 ~ dy ~ dx = \int_0^2 \int_u^{\frac u 2 + 1} \frac{1}{\sqrt u} ~ dv ~ du$
But say, you were integrating $f(x) = x$ over the given region. Due to symmetry of the original region about $x = 0$ and given $f(x) = x$ is an odd function,
$ \displaystyle \int_{- \sqrt 2}^{\sqrt 2} \int_{x^2}^{\frac {x^2} 2 + 1} x ~ dy ~ dx = 0$
But after change of variable, when $x \lt 0$, the integrand is $ - \sqrt u$ and when $x \gt 0$, the integrand is $ \sqrt u$.
So, $I = I_1 + I_2 = 0$ where,
$ \displaystyle I_1 = \int_0^2 \int_u^{\frac u 2 + 1} -\frac{1}{2} ~ dv ~ du$
$ \displaystyle I_2 = \int_0^2 \int_u^{\frac u 2 + 1} \frac{1}{2} ~ dv ~ du$
