Arc Length and Surface Area Revisited — Question 2

Question 2 – Solution

Solution

1. Choose a representation and interval that trace the desired curve exactly once. The equivalent arc-length formulas are $$\begin{aligned} L&=\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx,\\ L&=\int\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}\,dt,\\ L&=\int\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta. \end{aligned}$$

2. For a surface of revolution, multiply the appropriate arc-length element by $2\pi$ times the nonnegative distance to the axis. Check the tracing interval to prevent geometric double-counting.

3. The parametric speed is $3$ on $0\le t\le2\pi$. The polar arc-length factor is $6$ on $-\pi/2\le\theta\le\pi/2$. Each interval traces the circle once, so both forms give $L=6\pi$.

4. The best choice is the one whose single-tracing interval is clearest.