I apologize for the outage on the site yesterday and today. Lamar University is in Beaumont Texas and Hurricane Laura came through here and caused a brief power outage at Lamar. Things should be up and running at this point and (hopefully) will stay that way, at least until the next hurricane comes through here which seems to happen about once every 10-15 years. Note that I wouldn't be too suprised if there are brief outages over the next couple of days as they work to get everything back up and running properly. I apologize for the inconvienence.
Paul
August 27, 2020
Section 3-10 : Surface Area with Polar Coordinates
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 the tangent and arc length sections we’ll write the curve in terms of a set of parametric equations.
\[\begin{align*}x = r\cos \theta \hspace{0.75in}y = r\sin \theta \\ & \,\,\,\, = f\left( \theta \right)\cos \theta \hspace{0.75in} = f\left( \theta \right)\sin \theta \end{align*}\]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.