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    Home / Calculus II / Parametric Equations and Polar Coordinates / Arc Length and Surface Area Revisited
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    August 27, 2020

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    Section 3-11 : Arc Length and Surface Area Revisited

    We won’t be working any examples in this section. This section is here solely for the purpose of summarizing up all the arc length and surface area problems.

    Over the course of the last two chapters the topic of arc length and surface area has arisen many times and each time we got a new formula out of the mix. Students often get a little overwhelmed with all the formulas.

    However, there really aren’t as many formulas as it might seem at first glance. There is exactly one arc length formula and exactly two surface area formulas. These are,

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

    The problems arise because we have quite a few \(ds\)’s that we can use. Again, students often have trouble deciding which one to use. The examples/problems usually suggest the correct one to use however. Here is a complete listing of all the \(ds\)’s that we’ve seen and when they are used.

    \[\begin{align*}ds & = \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \,dx & \hspace{0.5in}& {\mbox{if }}y = f\left( x \right),\,\,a \le x \le b\\ ds & = \sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} \,dy & \hspace{0.5in} & {\mbox{if }}x = h\left( y \right),\,\,c \le y \le d\\ ds & = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} \,\,dt & \hspace{0.5in} & {\mbox{if }}x = f\left( t \right),y = g\left( t \right),\,\,\alpha \le t \le \beta \\ ds & = \sqrt {{r^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} \,d\theta & \hspace{0.5in} & {\mbox{if }}r = f\left( \theta \right),\,\,\alpha \le \theta \le \beta \end{align*}\]

    Depending on the form of the function we can quickly tell which \(ds\) to use.

    There is only one other thing to worry about in terms of the surface area formula. The \(ds\) will introduce a new differential to the integral. Before integrating make sure all the variables are in terms of this new differential. For example, if we have parametric equations we’ll use the third \(ds\) and then we’ll need to make sure and substitute for the x or y depending on which axis we rotate about to get everything in terms of t.

    Likewise, if we have a function in the form \(x = h\left( y \right)\) then we’ll use the second \(ds\) and if the rotation is about the y-axis we’ll need to substitute for the x in the integral. On the other hand, if we rotate about the x-axis we won’t need to do a substitution for the y.

    Keep these rules in mind and you’ll always be able to determine which formula to use and how to correctly do the integral.

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