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 ${\tfrac {\text{length of sector}}{\text{radius}}}$

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

Combining what we know, the surface area formula is:

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

• $l$ 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

${\begin{array}{rcl}{\tfrac {l_{1}}{r_{1}}}&=&{\tfrac {l_{1}+1}{r_{2}}}\\r_{2}l_{1}&=&r_{1}l_{1}+r_{1}l\\r_{1}l&=&(r_{2}-r_{1})l_{1}\\\end{array}}$

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:

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

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

$SA=2\pi \int _{a}^{b}{r\,ds}$ ...which then becomes...
$SA_{\text{x-axis}}=2\pi \int _{a}^{b}{y{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}\,dx}$	$SA_{\text{y-axis}}=2\pi \int _{a}^{b}{x{\sqrt {1+\left({\frac {dx}{dy}}\right)^{2}}}\,dy}$

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 $y$: $ds={\sqrt {1+\left({\frac {dx}{dy}}\right)}}\,dy$

$S=2\pi rl$

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

${\begin{aligned}SA&=2\pi \int _{0}^{1}{{\sqrt {y}}{\sqrt {\left({\frac {1}{2{\sqrt {y}}}}\right)^{2}+1}}\,dy}\\&=2\pi \int _{0}^{1}{{\sqrt {{\frac {1}{4}}+y}}\,dy}\\&={\frac {4\pi }{3}}\left({\frac {1}{4}}+y\right)^{3/2}{\bigg |}_{0}^{1}\\&={\frac {4\pi }{3}}\left(\left({\frac {5}{4}}\right)^{3/2}-\left({\frac {1}{4}}\right)^{3/2}\right)\end{aligned}}$

Method 2

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

$S=2\pi rl$

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

${\begin{aligned}S=&\int _{0}^{1}{2\pi x{\sqrt {1+4x^{2}}}\,dx}\\=&{\frac {\pi }{4}}\,{\frac {2}{3}}\,\left(1+4x^{2}\right)^{3/2}{\bigg |}_{0}^{1}\\=&{\frac {\pi }{6}}\,\left(5^{3/2}-1\right)\end{aligned}}$

Example 2

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

${\begin{aligned}S=&\int _{0}^{2\pi }{2\pi r\,ds}\\=&\int _{0}^{2\pi }{2\pi \left(1+\sin {t}\right){\sqrt {\sin ^{2}{t}+\cos ^{2}{t}}}\,dt}\\=&2\pi \int _{0}^{2\pi }{1+\sin {t}\,dt}\\=&2\pi \left[t-\cos {t}\right]_{0}^{2\pi }\\=&2\pi \left(2\pi -\cos {2\pi }-0+\cos {0}\right)\\=&4\pi ^{2}\end{aligned}}$