Suppose we have an strangely-shaped region \(S\), a mapping \(\Phi\), and a nicely-shaped region \(R\) such that \(\Phi(R) = S\). This is the ideal scenario for applying the change of variables theorem in order to integrate \(S\) (with the technical requirements that \(R\) is compact, \(\Phi \in C^1\), and \(\Phi\) is injective on \(R^0\)). Generally, you will be given \(S\) and you will need to come up with \(\Phi\) and \(R\).
In this scenario, change of variables tells us that
$$ \int_S f = \int_R \Phi \circ f \cdot | \Phi^\prime | $$for \( f : S \rightarrow \mathbb{R} \) integrable.
Below on the right we have our target green area \(S\) and on the left is the green \(R\). We know that \( \Phi(R) = S \). In this case we can take \( \Phi \) to be the polar mapping \( (r,\theta) \mapsto (r \cos \theta, r \sin \theta) \). Drag around the blue box on the left, which represents \(\text{d}R\) (a bit of \(R\)). The corresponding \( \Phi(\text{d}R) \) is the blue area on the right, and the overlaying red area is \( \Phi^\prime(\text{d}R) + \Phi(x) \), where \(x\) is the lower right corner of \(\text{d}R\). Note how different positions of \(\text{d}R\) affect the orientation and scale of the areas on the right. Additionally, you can scale \(\text{d}R\) with the red circle at its top-right corner.
Again, we have another very similar setup except that here \(\Phi\) is the mapping \( (u,v) \mapsto ( u^2-v^2, 2 u v ) \), which represents the complex squaring map under the usual identification \(\mathbb{C} \cong \mathbb{R}^2\). Drag around and scale \(\text{d}R\) and note how its different positions and sizes affect the orientation and scale of \( \Phi(\text{d}R) \) and \( \Phi^\prime(\text{d}R) + \Phi(x) \).