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
We will be looking at surface area in polar coordinates in this section. Note however that all we’re going to do is give the formulas for the surface area since most of these integrals tend to be fairly difficult.
We want to find the surface area of the region found by rotating,
\[r = f\left( \theta \right)\hspace{0.25in}\hspace{0.25in}\alpha \le \theta \le \beta \]about the \(x\) or \(y\)-axis.
As we did in thetangent andarc length sections we’ll write the curve in terms of a set of parametric equations.
\[\begin{align*}x & = r\cos \theta \hspace{0.75in}y = r\sin \theta \\ & = f\left( \theta \right)\cos \theta \hspace{0.55in} = f\left( \theta \right)\sin \theta \end{align*}\]If we now use the parametric formula for finding the surface area we’ll get,
where,
\[ds = \sqrt {{r^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} \,d\theta \hspace{0.25in}\hspace{0.25in}r = f\left( \theta \right),\,\,\,\,\,\alpha \le \theta \le \beta \]Note that because we will pick up a \(d\theta \) from the \(ds\) we’ll need to substitute one of the parametric equations in for \(x\) or \(y\) depending on the axis of rotation. This will often mean that the integrals will be somewhat unpleasant and so we will not be doing an example in this section.