I am wondering how I would find the partial derivative of $z = g(r, \theta) = \theta$ with respect to both $r$ and $\theta$. I realize that if you take the partial in respect to $\theta$, it is 1. I'm not sure how to take the partial of a function that does not have the included variable.
The full question is:
If $z = g(r, \theta)$ is the equation in cylindrical coordinates of a surface lying above a region $D$ in the $xy$-plane, then the area of the surface is given by the formula: $$ \text{surface area} = \iint \left(1 + \left(\frac{\partial z}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial z}{\partial\theta}\right)^2\right)^{1/2} r \, dr\, d\theta $$
Find the area of the “spiral ramp” $z = g(r, \theta) = \theta$ over the region $D$: $0 \leq r \leq 1$, and $0 \leq \theta \leq \pi$.
Think about the definition of the derivative:
$$\frac{\partial g}{\partial r}(r,\theta) = \lim_{\Delta r\to0}\frac{g(r+\Delta r,\theta)-g(r,\theta)}{\Delta r} = \lim_{\Delta r\to0}\frac{\theta - \theta}{\Delta r} = \lim_{\Delta r\to0}\frac{0}{\Delta r}= 0$$
The same is true any time $g(r,\theta)$ only depends on $\theta$.