Find the area of rectangle bounded by $u=0,~ u=a,~ v=0, ~v=1$ on helicoid defined by $x = u\cos v, ~ y=u\sin v, ~ z = av$.
I have solved many problems of similar kind in my calculus course, but this problem is from differential geometry book which I'm self-studying. I don't have any algorithm in my mind. I can't imagine the graph too. Maybe some kind of variable-change to flatten the surface.
Any help is appreciated.
A surface $S$ parameterized by a vector-valued function $\vec r(u,v)$ has area equal to the surface integral
$$\iint_S\mathrm dA=\int_0^1\int_0^a\left\|\frac{\partial\vec r}{\partial u}\times\frac{\partial\vec r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$$
You're generously provided with the parameterization and domain of integration. Differentiating $\vec r(u,v)$ gives
$$\vec r(u,v)=\left\langle u\cos v,u\sin v,av\right\rangle\implies\begin{cases}\frac{\partial\vec r}{\partial u}=\left\langle\cos v,\sin v,0\right\rangle\\\frac{\partial\vec r}{\partial v}=\left\langle-u\sin v,u\cos v,a\right\rangle\end{cases}$$
Find the normal vector to $S$, which is the cross product of these partial derivatives:
$$\frac{\partial\vec r}{\partial u}\times\frac{\partial\vec r}{\partial v}=\left\langle a\sin v,-a\cos v,u\right\rangle$$
Compute its norm:
$$\left\|\frac{\partial\vec r}{\partial u}\times\frac{\partial\vec r}{\partial v}\right\|=\sqrt{(a\sin v)^2+(-a\cos v)^2+u^2}=\sqrt{a^2+u^2}$$
You can think of this norm factor as an extension of the Jacobian determinant you typically use for integrating over a plane. Roughly speaking, it essentially scales up an area over a flat plane to compensate for the curvature of the surface. (I think the linked Wikipedia article does a decent job of explaining this more rigorously, and in fact that's where I first picked up the concept!)
So the area would be
$$\int_0^1\int_0^a\sqrt{a^2+u^2}\,\mathrm du\,\mathrm dv=\int_0^a\sqrt{a^2+u^2}\,\mathrm du$$
and the remaining integral is elementary.