Finding the Volume Enclosed by the Complex Plane and the Magnitude of a Complex Function

Question

How would one find the volume of the region under the magnitude of a complex function over a given region of the complex plane? For example for the complex function: $$f(z) = z$$ for $z \in \Bbb{C} $, how to calculate the volume under $|f(z)|$ for the square region bounded by $0$ and $1$ on the real axis and bounded by $0$ and $i$ on the imaginary axis?

Answer 1

Polar coordinates are well adapted to express $|f(z)|=|z|=r$ but a square has $\cos$ and $\sin$ at denominator which is not so nice to integrate.

Converting unit square domain in (x,y) to polar coordinates

As you can see in the answer above, it is feasible, but already complicated for even simple functions.

For $f(z)=z$ this gives $\displaystyle I=\int_0^{\frac\pi4}\int_0^{1/\cos(\theta)}r^2drd\theta+\int_{\frac\pi4}^{\frac\pi2}\int_0^{1/\sin(\theta)}r^2drd\theta$

And you can conclude with : Integral $ \int \frac{\operatorname d\!x}{\sin^3 x} $

Result: $I=\frac 16(2\sqrt{2}+\operatorname{argch}(3))$

In some cases, it may be easier to keep integrating on the square domain $\sqrt{x^2+y^2}$ where $x(z)=\Re(f(z))$ and $y(z)=\Im(f(z))$ as suggested by the accepted answer to the first referenced thread, but I think it all depends of what $f(z)$ looks like.

$\displaystyle I=\int_0^1\int_0^1 \sqrt{x^2+y^2}dxdy$ and I guess a change of variable like $x=y\sinh(u)$ lead to the same kind of calculation than previously.

The result $I\simeq0.7652$ is slightly greater than $\frac 18(\frac 43\pi R^3)=\frac\pi6\simeq 0.5236$ which is the volume of the first octant of the unit sphere, which is reassuring.