Area of a rectangle under transformation $e^{z}$ Given that $$\Omega=\left\{x+ \iota y:-1 \leq x \leq2,\frac{-\pi}{3}\leq y \leq \frac{\pi}{3}\right\}$$ 
Now question is what will be the area of this rectangle under transforamtion
$$x+\iota y \to e^{x+ \iota y}$$
Solution i tried -The given Transformation is a rotation so it will only rotate the  given curve ,but i can find the proper solution i.e proper area of transformed curve .
please help!
 A: Since the area of rectangle $\Omega$ can be gotten through
$$\int_\Omega 1 \ dz = \int_{-1}^2 \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}1 \ dy \ dx = (2-(-1))\cdot\left(\frac{\pi}{3}-\left(-\frac{\pi}{3}\right)\right) = 2\pi,$$
the transformed integral is simply:
$$\int_\Omega e^z \ dz = \int_{-1}^2 \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} e^{x+iy} \ dy \ dx = \int_{-1}^2 \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} e^x(\cos(y)+i\sin(y)) \ dy \ dx = \\ = \int_{-1}^2 \left[e^x(\sin(y)-i\cos(y))\right]_{y=-\frac{\pi}{3}}^{y=\frac{\pi}{3}} \ dx = \int_{-1}^2 e^x\left(\frac{\sqrt{3}}{2}-\frac{i}{2}+\frac{\sqrt{3}}{2}-\frac{i}{2}\right) dx = \\ = \int_{-1}^2 \sqrt{3}\ e^x \  dx = \sqrt{3} \ \left[e^x\right]_{x=-1}^{x=2} = \sqrt{3}\ (e^2-e^{-1}).$$
Wolfram Alpha also shows the integral (in a slightly different form, since $e^2-e^{-1} = \frac{e^3-1}{e}$).
Edit: Your transformation is not a simple rotation, it's a combination of a rotation and an enlargement centered at the origin. The resulting shape will be a skewed rectangle with curved sides, that is "wider", the farther we move away from the origin. See more examples of it  on Wikipedia.
