MATH 152 Chapter 9.4

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Surface Area of Revolution

Abstraction 1: Surface Area around Cylinder

$SA=2\pi rh$

• $r$ is radius of base
• $h$ is the height of the cylinder.

Abstraction 2: Surface Area around Cone

When rolled out, cone's surface area becomes a sector of a circle whose radius is the slant height and whose length is the circumference of the base ($C=2\pi r$).

Recall that the central angle (in radians) of any sector is $\frac{\text{length of sector}}{\text{radius}}$

Also recall that the area of a sector is $\frac{1}{2}{r}^2\theta$, where $r$ is the radius and $\theta$ is the central angle.

Combining what we know, the surface area formula is:

$\begin{array}{rl}SA=& \frac{1}{2}{l}^2\theta \\ =& \frac{1}{2}{l}^2\left(\frac{2\pi r}{l}\right)\\ SA=& \pi rl\end{array}$

• $l$ is the slant hight
• $r$ is the radius of the base

Abstraction 3: Surface Area around Frustum

Frustum of cone with original slant height $l_2$ and radius $r_2$

Consult the figure to the right for better visualization.

Surface area of frustum is basically the difference between the larger and smaller surface area: $\pi r_2{l}_2-\pi r_1{l}_1$

When put in terms of $l$ (note that $l$ is the frustum slant height), the equation becomes $\pi r_2(l_1+l)-\pi r_1{l}_1$

Using similar triangles, the relationship between the slant heights and radii of the two cones is

$\begin{array}{rcl}\frac{l_1}{r_1}& =& \frac{l_1+1}{r_2}\\ r_2{l}_1& =& r_1{l}_1+r_{1}l\\ r_{1}l& =& (r_2-r_1)l_1\\ \end{array}$

Substitute this into the previous equation: $\pi (r_{1}l+r_{2}l)=\pi l(r_1+r_2)$

To put this in terms of a single radius, we can find the average:

$\begin{array}{rcl}r& =& \frac{1}{2}(r_{1}l+r_{2}l)\\ (r_1+r_2)& =& 2r\end{array}$

Plug the average radius into the previous equation, and bingo!

$SA=2\pi rl$

• $l$ is the slant height of the frustum
• $r$ is the average radius

Think of this as the average circumference × slant height, and our formula looks very similar to the cylinder formula (but they are not the same).

Surface Area of Revolution

When rotated about the $x$-axis, $f(x)$ becomes the radius. We must use arc length for the slant height $l$ because $dx$ represents the "plain height" between $r_1$ and $r_2$, not the "slant height". Here's why.

• $r$ becomes $x$ or $y$
• $l$ becomes arc length of $x+dx$ or $y+dy$

When we plug these new values into our equation $SA=2\pi rl$, we get

$SA=2\pi \underset{a}{\overset{b}{\int }}rds$ ...which then becomes...
$S{A}_{\text{x-axis}}=2\pi \underset{a}{\overset{b}{\int }}y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$	$S{A}_{\text{y-axis}}=2\pi \underset{a}{\overset{b}{\int }}x\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}dy$

Warm Up

Find area of the surface obtained by rotating the curve $y=x^2,x\in [0,1]$ about the $y$-axis.

$x=\sqrt{y}$

Method 1

Find surface area with respect to $y$: $ds=\sqrt{1+\left(\frac{dx}{dy}\right)}dy$

$S=2\pi rl$

• $r=x=\sqrt{y}$
• $l=ds=\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}dy$

$\begin{array}{rl}SA& =2\pi \underset{0}{\overset{1}{\int }}\sqrt{y}\sqrt{{\left(\frac{1}{2\sqrt{y}}\right)}^2+1}dy\\ & =2\pi \underset{0}{\overset{1}{\int }}\sqrt{\frac{1}{4}+y}dy\\ & =\frac{4\pi }{3}{(\frac{1}{4}+y)}^{3/2}{|}_0^1\\ & =\frac{4\pi }{3}({\left(\frac{5}{4}\right)}^{3/2}-{\left(\frac{1}{4}\right)}^{3/2})\end{array}$

Method 2

Find surface area with respect to $x$: $ds=\sqrt{1+\left(\frac{dy}{dx}\right)}dx$

$S=2\pi rl$

• $r=x$
• $l=\sqrt{a+{\left(\frac{dy}{dx}\right)}^2}dx$

$\begin{array}{rl}S=& \underset{0}{\overset{1}{\int }}2\pi x\sqrt{1+4x^2}dx\\ =& \frac{\pi }{4}\frac{2}{3}{(1+4x^2)}^{3/2}{|}_0^1\\ =& \frac{\pi }{6}(5^{3/2}-1)\end{array}$

Example 2

The circle $x=\cos t,y=1+\sin t$ is rotated about the $x$-axis. Find the area of the resulting surface.

$\begin{array}{rl}S=& \underset{0}{\overset{2\pi }{\int }}2\pi rds\\ =& \underset{0}{\overset{2\pi }{\int }}2\pi (1+\sin t)\sqrt{{\sin }^{2}t+{\cos }^{2}t}dt\\ =& 2\pi \underset{0}{\overset{2\pi }{\int }}1+\sin tdt\\ =& 2\pi {[t-\cos t]}_0^{2\pi }\\ =& 2\pi (2\pi -\cos 2\pi -0+\cos 0)\\ =& 4{\pi }^2\end{array}$