17Calculus Polar Coordinates - Surface Area
17Calculus
On this page we cover a common calculus problem involving polar coordinates, determining arc length.
As mentioned on the main polar coordinates page, polar coordinates are just parametric equations. If you are familiar with parametric equations, this material should be very intuitive.
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To calculate the surface area of a polar curve revolved about an axis, we use these integrals.
The equation of the polar curve is in the form \( x=r(\theta) \cos(\theta) \) and \( y = r(\theta) \sin(\theta) \) and the curve that is being revolved is defined to be from \(\theta_0\) to \(\theta_1\). We also define \( ds = \sqrt{r^2 + [dr/d\theta]^2}~dt \).
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rotation about the x-axis |
\(\displaystyle{ S = 2\pi \int_{\theta_0}^{\theta_1}{y~ds} = }\) \(\displaystyle{ 2\pi \int_{\theta_0}^{\theta_1}{ r(\theta) \sin(\theta) \sqrt{r^2 + [dr/d\theta]^2}~dt } }\) | |
rotation about the y-axis |
\(\displaystyle{ S = 2\pi \int_{\theta_0}^{\theta_1}{x~ds} = }\) \(\displaystyle{ 2\pi \int_{\theta_0}^{\theta_1}{ r(\theta) \cos(\theta) \sqrt{r^2 + [dr/d\theta]^2}~dt } }\) | |
So, why the \(ds\) term? If you look at the previous page on arc length, you will notice that this term appears in that integral. This is called a differential length and is just a convenient way of writing these integrals.
Practice
Calculate the surface area formed by rotating the polar curve \(r=4\sin\theta\), \(0\leq\theta\leq\pi\) about the polar axis.
Problem Statement
Calculate the surface area formed by rotating the polar curve \(r=4\sin\theta\), \(0\leq\theta\leq\pi\) about the polar axis.
Solution
Here are two videos. The first one solves the given problem. The second one is almost identical except that the polar curve is \(r=\sin\theta\). It will help to watch both videos. Of course, the answer in the first video is four times the answer in the second.
Krista King Math - 1268 video solution
video by Krista King Math |
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MIP4U - 1268 video solution
video by MIP4U |
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