Calculus II - Surface Area with Polar Coordinates

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Section 9.10 : 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.

$\begin{align*}x & = r\cos \theta \hspace{0.75in}y = r\sin \theta \\ & = f\left( \theta \right)\cos \theta \hspace{0.55in} = f\left( \theta \right)\sin \theta \end{align*}$

If we now use the parametric formula for finding the surface area we’ll get,

$\begin{align*}S & = \int{{2\pi y\,ds}}\hspace{0.5in}{\mbox{rotation about }}x - {\rm{axis}}\\ S & = \int{{2\pi x\,ds}}\hspace{0.5in}{\mbox{rotation about }}y - {\rm{axis}}\end{align*}$

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.