For this question I have managed to sketch the region however I don't understand how to solve this sort of a problem where we have two regions? Also when we use the change of variable formula with the Jacobian is that for one region only?
Here we have two regions. one where x>0 and one where x<0 but that's all that I have been able to understand from this question and would be grateful if someone could provide some explanation with how to approach this problem.
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$\begingroup$Since you sketched the regions in the $ \ xy-$ plane over which we wish to integrate, you likely found something that looks like this:
I have marked the intersection points between the pairs of semi-circles and pairs of parabolas; I haven't bothered to give the coordinates of those points, since we won't be using them anywhere. Notice the interesting feature that the "upper" parabola intersects the "lower" semi-circle at only one point.
Now, we are given a specific variable substitution to use: $ \ u \ = \ x^2 \ + \ y^2 \ , \ v \ = \ y \ - \ x^2 \ $ . This has the effect of turning the pairs of curves into
$$ y \ = \ \sqrt{1 \ - \ x^2} \ \ \rightarrow \ \ u \ = \ 1 \ \ , \ \ y \ = \ \sqrt{4 \ - \ x^2} \ \ \rightarrow \ \ u \ = \ 4 \ \ ; $$
$$ y \ = \ x^2 \ \ \rightarrow \ \ v \ = \ 0 \ \ , \ \ y \ = \ 1 \ + \ x^2 \ \ \rightarrow \ \ v \ = \ 1 \ \ . $$
We now have two pairs of parallel lines forming a single rectangle with just four intersection points:
Our original two regions now "overlap" onto one rectangle. If we work out what becomes of the intersection points, we find that the former "right-hand" region is "mapped" onto the rectangle with the "circulation" of the intersections $ \ A'B'C'D' \ $ being counter-clockwise ("positive" orientation). However, the former "left-hand" region is mapped onto the same rectangle, but with the path through intersections $ \ D'E'F'G' \ $ being clockwise ("negative" orientation).
This may be dropping us a hint about what the value of the integral is. If we were calculating an integral of some function of a single variable in one direction over a particular interval, and added it to an integral of the same function over the same interval in the opposite direction, what would happen?
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The Jacobian determinant of the transformation is given by
$$ \mathfrak{J} \ = \ \frac{1}{\left| \begin{array}{cc}\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}\\\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}\end{array}\right|} \ = \ \frac{1}{\left| \begin{array}{cc} 2x&2y\\-2x&1\end{array}\right|} \ = \ \frac{1}{2x \ (1+2y)} \ \ . $$
Notice that the Jacobian becomes undefined for $ \ x \ = \ 0 \ $ , that is, just at the point $ \ D \ $ . It does apply to both regions, but has opposite signs on either side of the $ \ y-$ axis. (It also provides a valuable simplification in transforming the integral you are asked to compute.)
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