Surface area given a function in polar without converting to rectangular

Question

Note: I need to use the double integral method of integration for this.

Let's say I have a function in terms of $r$ and $θ$: $f(r, θ)$

For a rectangular function $f(x, y)$, one can use the following formula to find the surface area:

$$\int\int_D\left(\sqrt{f_x^2+f_y^2+1}\right)dA$$

Is there a formula for doing it with a function of $r$ and $\theta$?

Update:

Let's say we have the function $f(r, θ)=9-r^2$ and have the restriction $2≤r≤\sqrt{6}$.

So, it would go from this to this.

So, to answer the question of what I am looking for, let's say I want the surface area of the function, I would want the surface area of the above function with the above restriction. It could probably somehow be done using the same reasoning that one uses the substitution rule, but I do not know how to do this.

Answer 1

If your surface $S$ is given in the form $z=f(r,\theta)$, where $r$ and $\theta$ are polar coordinates in the $(x,y)$-plane, then this is equivalent to a parametric representation $$(r,\theta)\mapsto {\bf r}(r,\theta)=\bigl(r\cos \theta,r\sin\theta, f(r,\theta)\bigr)$$ with a domain $D$ to be specified by inequalities referring to $r$ and $\theta$. For the area element on $S$ we have to compute $${\bf r}_r=(\cos\theta,\sin\theta, f_r\bigr),\qquad {\bf r}_\theta=\bigl(-r\sin\theta,r\cos\theta,f_\theta\bigr)$$ and then $${\bf r}_r\times{\bf r}_\theta=\bigl(f_\theta\sin\theta -rf_r\cos\theta ,-r f_r\sin\theta-f_\theta\cos\theta,r\bigr)\ ,$$ so that $$|{\bf r}_r\times{\bf r}_\theta|=\sqrt{r^2(1+ f_r^2)+f^2_\theta}\ .$$It follows that the area of $S$ is given by $${\rm area}(S)=\int_D\sqrt{r^2(1+ f_r^2)+f^2_\theta}\>{\rm d}(r,\theta)\ .$$

Answer 2

If the function is already in polar coordinates you don't need to do the substitution, it's already in the right form. You will still need to integrate $f(r,\theta)r$ instead of $f(r,\theta)$ to calculate the correct area as though you had done the substitution. This is because we typically define area with regards to rectangular coordinates and need to convert back to calculate all of the approximating rectangles when setting up the integral even though the function was never in the form $f(x,y)$.