Parametrizing the surface $z=\log(x^2+y^2)$

Question

Let $S$ be the surface given by $$z = \log(x^2+y^2),$$ with $1\leq x^2+y^2\leq5$. Find the surface area of $S$.

I'm thinking the approach should be $$A(s) = \iint_D \ |\textbf{T}_u\times \textbf{T}_v| \,du\,dv ,$$ where $T_u, T_v$ are tangent vectors to the surface (and of course their cross product is the normal).

Now my issue is the parametrization bit. I'm having a very hard time grasping the concept of parametrization of surfaces. I'd be really grateful if someone can help with parametrizing the surface, i.e. $x(u,v)$, $y(u,v)$ and $z(u,v)$ and the domains of $u$ and $v$.

Answer 1

When you use the cross product method, the Jacobian takes care of itself. Along the surface the $\vec r=\langle x,y,z\rangle=\langle r\cos\theta,r\sin\theta,2\ln r\rangle$. Then the total differential along the surface is $$d\vec r=\langle\cos\theta,\sin\theta,\frac2r\rangle\,dr+\langle-r\sin\theta,r\cos\theta,0\rangle\,d\theta$$ Then we can get the vector areal element via the cross product $$\begin{align}d^2\vec A&=\pm\langle\cos\theta,\sin\theta,\frac2r\rangle\,dr\times\langle-r\sin\theta,r\cos\theta,0\rangle\,d\theta\\ &=\pm\langle-2\cos\theta,-2\sin\theta,r\rangle\,dr\,d\theta\end{align}$$ Then the scalar areal element is $$d^2A=\left|\left|d^2\vec A\right|\right|=\sqrt{4+r^2}\,dr\,d\theta$$ Then $$A=\int d^2A=\int_0^{2\pi}\int_1^\sqrt5\sqrt{4+r^2}\,dr\,d\theta$$ In the inner integral, let $r=2\sinh u$. Then $$\begin{align}\int\sqrt{4+4^2}\,dr&=\int2^2\cosh^2u\,du=2\int(\cosh2u+1)du=2\left(\frac12\sinh2u+u\right)+C_1\\ &=2\left(\sinh u\cosh u+u\right)+C_1\\ &=2\left(\frac r2\sqrt{\frac{r^2}{4}+1}+\ln\left(\frac r2+\sqrt{\frac{r^2}4+1}\right)\right)+C_1\\ &=\frac12r\sqrt{4+r^2}+2\ln\left(r+\sqrt{4+r^2}\right)+C\end{align}$$ So $$\begin{align}A&=2\pi\left[\frac12r\sqrt{4+r^2}+2\ln\left(r+\sqrt{4+r^4}\right)\right]_1^{\sqrt5}\\ &=2\pi\left[\frac123\sqrt5+2\ln\left(\sqrt5+3\right)-\frac12\sqrt5-2\ln2\right]=2\pi\left[\sqrt5+4\ln\left(\frac{\sqrt5+1}2\right)\right]\end{align}$$ You just knew the golden ratio was coming when you saw the $\sqrt5$ in the limits!

Answer 2

It looks like polar would be good for this, i.e. $x=u\cos v,\ y=u \sin v$ where $1\le u \le 5$ and $0 \le v \le 2\pi.$ [The last should technically have $v<2\pi$ but that doesn't matter in integration.] Then $z$ from your formula. $z=\log u^2$

Oops it's $1 \le u \le \sqrt{5}.$

Answer 3

If a geometric object has a particular symmetry, generally parameterizations that reflect that symmetry result in easier computations. In our case, $S$ is the surface of a graph whose domain $A := \{1 \leq x^2 + y^2 \leq 5\}$ is an annulus centered at the origin. This suggests letting one of our parameterization variables be the distance $r$ of a point $p \in A$ to the origin, that set, setting $$r := \sqrt{x^2 + y^2}.$$ Then, we can see that our domain $A$ takes on an especially simple form: $$A := \{1 \leq r \leq \sqrt{5}\} .$$ When we use this $r$ for one of our parameters, usually we take an angular parameter $\theta$ for the other (the coordinate system $(r, \theta)$ is called polar coordinates). Since our region sweeps all the way around our origin, we should let $\theta$ vary over some (half-open) interval of length $2 \pi$, e.g., $[0, 2 \pi]$.

Drawing a picture, we see using elementary trigonometry that the transformation that expresses $x, y$ in terms of our polar coordinates $r, \theta$ is $$\left\{\begin{array}{rcl}x(r, \theta) & = & r \cos \theta \\ y(r, \theta) & = & r \sin \theta\end{array}\right.$$

Answer 4

One technique that I found useful when I was teaching calc III this past semester is to parametrize a surface of the form $z=f(x,y)$ as $\mathbf{r}(x,y)=(x,y,f(x,y))$, get a double integral over some region, and then change coordinates in that integral. This is an alternative to directly choosing your parametrization in the "convenient" coordinates (in this case polar coordinates). One nice thing about this is that $\mathbf{r}_x \times \mathbf{r}_y = (-f_x,-f_y,1)$, so if you remember that, you can avoid doing a cross product. By contrast if you actually parametrize by $r,\theta$, taking $\mathbf{r}(r,\theta)=(r\cos(\theta),r\sin(\theta),f(r\cos(\theta),r\sin(\theta)))$, computing $\mathbf{r}_r \times \mathbf{r}_\theta$ is slightly annoying.

One additional point to pay attention to: when you change coordinates from this Cartesian integral I mentioned to polar, you need the polar Jacobian (the extra factor of $r$). When you parametrize directly in polar coordinates, you do not need the polar Jacobian; this factor of $r$ appears inside $\mathbf{r}_r \times \mathbf{r}_\theta$.