Surface integral in complex function

Question

The question:Let $\mathbb{C^2}$={($z$,$w$)|$z$ and $w$ are complex number}.Find the area of the part of the surface $w=z^2$,where$|z|\le1$.
My thought:Viewing $\mathbb{C}$ as $\mathbb{R^2}$, $|z|\le1$ can be parametrized by polar coordinates $(r,\theta)$.
Then the above map $w=z^2$ can be viewed as a map from $\mathbb{R^2}$to $\mathbb{R^2}$.Are we going to find volume of $\mathbb{R^2}$ in $\mathbb{R^4}$ by integration?Applying the formula $\int_S\sqrt{\det{D^TD}}$?I am confused.Could anyone help?

Answer 1

You are right. We (can; it's not the only way) find the surface area by parametrising it and integrating $\sqrt{\det D^TD}$ - where $D$ is the differential matrix of the parametrisation - over the parameter domain.

I would compute the derivative of the parametrisation in Euclidean coordinates and only then switch to polar coordinates to calculate the area. In complex form, the parametrisation is $z \mapsto (z,z^2)$, and the real form is then $(x,y) \mapsto (x,y,x^2-y^2,2xy)$, from which we obtain

$$D = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 2x & -2y \\ 2y & 2x \end{pmatrix}$$

and then the area as

$$\int\limits_{x^2+y^2 \leqslant 1} 1 + 4(x^2+y^2)\,dx\,dy = 2\pi \int_0^1 \bigl(1 + 4r^2\bigr)r\,dr = 3\pi.$$