Surface Area with Parametric Equations — Question 10

Question 10 – Solution

Solution

1. Let $r(s)$ denote the nonnegative distance from a point on a generating curve to the axis, where $s$ is arc length. Its arc-length-weighted average distance is $$r_{\mathrm{avg}}=\frac{1}{L}\int_Cr\,ds.$$

2. Multiply by $L$: $$\int_Cr\,ds=Lr_{\mathrm{avg}}.$$ The surface area of revolution is therefore $$S=2\pi\int_Cr\,ds=2\pi Lr_{\mathrm{avg}}.$$

3. If two curves have the same length $L$ and the same arc-length-weighted average radius, this formula gives the same value of $S$ for both.

4. Thus the answer is , provided each generating curve sweeps its surface exactly once. If a parametrization retraces the curve or distinct pieces generate the same surface, the integral counts multiplicity and the comparison requires removing that double-counting first.