I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
Section 9.9 : Arc Length with Polar Coordinates
For problems 1 – 3 determine the length of the given polar curve. For these problems you may assume that the curve traces out exactly once for the given range of \(\theta \).
- \(\displaystyle r = \frac{1}{{\cos \theta }}\), \(\displaystyle 0 \le \theta \le \frac{\pi }{3}\)
- \(r = {\theta ^2}\), \(0 \le \theta \le 3\pi \)
- \(r = 6\cos \theta - 3\sin \theta \), \(0 \le \theta \le \pi \)
For problems 4 – 6 set up, but do not evaluate, an integral that gives the length of the given polar curve. For these problems you may assume that the curve traces out exactly once for the given range of \(\theta \).
- \(r = \sin \left( {{\theta ^2}} \right)\), \(0 \le \theta \le \pi \)
- \(r = \cos \left( {1 + \sin \theta } \right)\), \(0 \le \theta \le 2\pi \)
- \(r = {{\bf{e}}^{ - \,\,\frac{1}{4}\theta }}\cos \theta \), \(0 \le \theta \le 3\pi \)