# Area of a Surface of Revolution

A surface of revolution is obtained when a curve is rotated about an axis.

We consider two cases - revolving about the $$x-$$axis and revolving about the $$y-$$axis.

## Revolving about the $$x-$$axis

Suppose that $$y\left( x \right),$$ $$y\left( t \right),$$ and $$y\left( \theta \right)$$ are smooth non-negative functions on the given interval.

1. If the curve $$y = f\left( x \right),$$ $$a \le x \le b$$ is rotated about the $$x-$$axis, then the surface area is given by
$A = 2\pi \int\limits_a^b {f\left( x \right)\sqrt {1 + {{\left[ {f^\prime\left( x \right)} \right]}^2}} dx} .$
2. If the curve is described by the function $$x = g\left( y \right),$$ $$c \le y \le d,$$ and rotated about the $$x-$$axis, then the area of the surface of revolution is given by
$A = 2\pi \int\limits_c^d {y\sqrt {1 + {{\left[ {g^\prime\left( y \right)} \right]}^2}} dy} .$
3. If the curve defined by the parametric equations $$x = x\left( t \right),$$ $$y = y\left( t \right),$$ with $$t$$ ranging over some interval $$\left[ {\alpha,\beta} \right],$$ is rotated about the $$x-$$axis, then the surface area is given by the following integral (provided that $$y\left( t \right)$$ is never negative)
$A = 2\pi \int\limits_\alpha ^\beta {y\left( t \right)\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .$
4. If the curve defined by polar equation $$r = r\left( \theta \right),$$ with $$\theta$$ ranging over some interval $$\left[ {\alpha,\beta} \right],$$ is rotated about the polar axis, then the area of the resulting surface is given by the following formula (provided that $$y = r\sin \theta$$ is never negative)
$A = 2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\sin \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .$

## Revolving about the $$y-$$axis

The functions $$g\left( y \right),$$ $$x\left( t \right),$$ and $$x\left( \theta \right)$$ are supposed to be smooth and non-negative on the given interval.

1. If the curve $$y = f\left( x \right),$$ $$a \le x \le b$$ is rotated about the $$y-$$axis, then the surface area is given by
$A = 2\pi \int\limits_a^b {x\sqrt {1 + {{\left[ {f^\prime\left( x \right)} \right]}^2}} dx} .$
2. If the curve is described by the function $$x = g\left( y \right),$$ $$c \le y \le d,$$ and rotated about the $$y-$$axis, then the area of the surface of revolution is given by
$A = 2\pi \int\limits_c^d {g\left( y \right)\sqrt {1 + {{\left[ {g^\prime\left( y \right)} \right]}^2}} dy} .$
3. If the curve defined by the parametric equations $$x = x\left( t \right),$$ $$y = y\left( t \right),$$ with $$t$$ ranging over some interval $$\left[ {\alpha,\beta} \right],$$ is rotated about the $$y-$$axis, then the surface area is given by the integral (provided that $$x\left( t \right)$$ is never negative)
$A = 2\pi \int\limits_\alpha ^\beta {x\left( t \right)\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .$
4. If the curve defined by polar equation $$r = r\left( \theta \right),$$ with $$\theta$$ ranging over some interval $$\left[ {\alpha,\beta} \right],$$ is rotated about the $$y-$$axis, then the area of the resulting surface is given by the formula (provided that $$x = r\cos \theta$$ is never negative)
$A = 2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\cos \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the lateral surface area of a right circular cone with slant height $$\ell$$ and base radius $$R.$$

### Example 2

The catenary line $y = a\cosh \frac{x}{a}$ rotates around the $$x-$$axis and sweeps out a surface called a catenoid. Find the surface area of the catenoid when $$x \in \left[ { - a,a} \right].$$

### Example 3

Find the area of the surface obtained by revolving the astroid $x = {\cos^3}t,\, y = {\sin^3}t$ around the $$x-$$axis.

### Example 4

The lemniscate of Bernoulli given by the equation ${r^2} = {a^2}\cos 2\theta$ rotates around the polar axis. Find the area of the resulting surface.

### Example 1.

Find the lateral surface area of a right circular cone with slant height $$\ell$$ and base radius $$R.$$

Solution.

Let the slant height $$\ell$$ be defined by the equation $$y = f\left( x \right) = kx.$$ The slope $$k$$ is given by

$k = \tan \alpha = \frac{R}{H},$

where $$H$$ is the height of the cone.

We calculate the lateral surface area of the cone by the formula

$A = 2\pi \int\limits_a^b {f\left( x \right)\sqrt {1 + {{\left[ {f^\prime\left( x \right)} \right]}^2}} dx} .$

Substituting

$a = 0,\;b = H,\;f\left( x \right) = kx = \frac{R}{H} x,\;f^\prime\left( x \right) = k = \frac{R}{H},$

we obtain

$A = 2\pi \int\limits_0^H {kx\sqrt {1 + {k^2}} dx} = 2\pi k\sqrt {1 + {k^2}} \int\limits_0^H {xdx} = 2\pi k\sqrt {1 + {k^2}} \left. {\frac{{{x^2}}}{2}} \right|_0^H = \pi k\sqrt {1 + {k^2}} {H^2} = \pi \frac{{R\cancel{H^2}\sqrt {{R^2} + {H^2}} }}{{\cancel{H^2}}} = \pi R\sqrt {{R^2} + {H^2}} .$

By the Pythagorean theorem, $$\sqrt {{R^2} + {H^2}} = \ell.$$ Hence,

$A = \pi R\ell.$

### Example 2.

The catenary line $y = a\cosh \frac{x}{a}$ rotates around the $$x-$$axis and sweeps out a surface called a catenoid. Find the surface area of the catenoid when $$x \in \left[ { - a,a} \right].$$

Solution.

We find the surface area through integration by the formula

$A = 2\pi \int\limits_a^b {f\left( x \right)\sqrt {1 + {{\left[ {f^\prime\left( x \right)} \right]}^2}} dx} .$

We integrate here from $$-a$$ to $$a.$$ As $$f\left( x \right) = a\cosh \frac{x}{a},$$ then

$f^\prime\left( x \right) = \left( {a\cosh \frac{x}{a}} \right)^\prime = a\sinh \frac{x}{a} \cdot \frac{1}{a} = \sinh \frac{x}{a}.$

So we have

$A = 2\pi \int\limits_{ - a}^a {a\cosh \frac{x}{a}\sqrt {1 + {{\left[ {\sinh \frac{x}{a}} \right]}^2}} dx} .$

Recall the following hyperbolic identities:

$1 + \left( {\sinh x} \right)^2 = \left( {\cosh x} \right)^2,\;\;\left( {\cosh x} \right)^2 = \frac{1}{2}\left[ {\cosh \left( {2x} \right) + 1} \right].$

This yields:

$A = 2\pi a\int\limits_{ - a}^a {{{\left( {\cosh \frac{x}{a}} \right)}^2}dx} = \pi a\int\limits_{ - a}^a {\left( {\cosh \frac{{2x}}{a} + 1} \right)dx} = \pi a\left. {\left[ {\frac{a}{2}\sinh \frac{{2x}}{a} + x} \right]} \right|_{ - a}^a = \pi a\left[ {\left( {\frac{a}{2}\sinh 2 + a} \right) - \left( {\frac{a}{2}\sinh \left( { - 2} \right) - a} \right)} \right] = \pi a\left( {a\sinh 2 + 2a} \right) = \pi {a^2}\left( {\sinh 2 + 2} \right).$

### Example 3.

Find the area of the surface obtained by revolving the astroid $x = {\cos^3}t,\, y = {\sin^3}t$ around the $$x-$$axis.

Solution.

When calculating the surface area, we consider the part of the astroid lying in the first quadrant and then multiply the result by $$2.$$ As the curve is defined in parametric form, we can write

$A = 2\pi \int\limits_a^b {y\left( t \right)\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} = 4\pi \int\limits_0^{\frac{\pi }{2}} {{{\sin }^3}t\sqrt {{{\left[ {\left( {{{\cos }^3}t} \right)^\prime} \right]}^2} + {{\left[ {\left( {{{\sin }^3}t} \right)^\prime} \right]}^2}} dt} .$

Find the derivatives:

$\left( {{{\cos }^3}t} \right)' = - 3\,{\cos ^2}t\sin t,\;\;\left( {{{\sin }^3}t} \right)' = 3\,{\sin ^2}t\cos t,$

$\left[ {\left( {{{\cos }^3}t} \right)'} \right]^2 + \left[ {\left( {{{\sin }^3}t} \right)'} \right]^2 = \left( { - 3\,{{\cos }^2}t\sin t} \right)^2 + \left( {3\,{{\sin }^2}t\cos t} \right)^2 = 9\,{\cos ^4}t\,{\sin ^2}t + 9\,{\sin ^4}t\,{\cos ^2}t = 9\,{\sin ^2}t\,{\cos ^2}t\underbrace {\left( {{{\cos }^2}t + {{\sin }^2}t} \right)}_{ = 1} = \left( {3\sin t\cos t} \right)^2.$

Hence, the surface area is

$A = 4\pi \int\limits_0^{\frac{\pi }{2}} {\left( {{{\sin }^3}t \cdot 3\sin t\cos t} \right)dt} = 12\pi \int\limits_0^{\frac{\pi }{2}} {{{\sin }^4}t\cos tdt} = 12\pi \cdot \left. {\frac{{{{\sin }^5}t}}{5}} \right|_0^{\frac{\pi }{2}} = \frac{{12\pi }}{5}.$

### Example 4.

The lemniscate of Bernoulli given by the equation ${r^2} = {a^2}\cos 2\theta$ rotates around the polar axis. Find the area of the resulting surface.

Solution.

We determine the surface area by the formula

$A = 2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\sin \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta .$

Due to symmetry, we can integrate from $$0$$ to $$\frac{\pi }{4}$$ considering the curve in the first quadrant and then multiply the result by $$2.$$ So

$A = 4\pi \int\limits_0^{\frac{\pi }{4}} {r\left( \theta \right)\sin \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2 + \left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta .$

Take the derivative:

$r^\prime\left( \theta \right) = \left( {a\sqrt {\cos 2\theta } } \right)^\prime = \frac{a}{{2\sqrt {\cos 2\theta } }} \cdot \left( { - \sin 2\theta } \right) \cdot 2 = - \frac{{a\sin 2\theta }}{{\sqrt {\cos 2\theta } }}.$

Hence, the derivative squared is written in the form

$\left[ {r^\prime\left( \theta \right)} \right]^2 = \left[ { - \frac{{a\sin 2\theta }}{{\sqrt {\cos 2\theta } }}} \right]^2 = \frac{{{a^2}{{\left( {\sin 2\theta } \right)}^2}}}{{\cos 2\theta }}.$

We can simplify the expression with the square root:

$\sqrt {{{\left[ {r\left( \theta \right)} \right]}^2 + \left[ {r^\prime\left( \theta \right)} \right]}^2} = \sqrt {{a^2}\cos 2\theta + \frac{{{a^2}{{\left( {\sin 2\theta } \right)}^2}}}{{\cos 2\theta }}} = a\sqrt {\frac{{{{\left( {\cos 2\theta } \right)}^2 + {\left( {\sin 2\theta } \right)}^2}}}{{\cos 2\theta }}} = \frac{a}{{\sqrt {\cos 2\theta } }}.$

Then

$A = 4\pi \int\limits_0^{\frac{\pi }{4}} {\left( {a\sqrt {\cos 2\theta } \sin \theta \cdot \frac{a}{{\sqrt {\cos 2\theta } }}} \right)d\theta } = 4\pi {a^2}\int\limits_0^{\frac{\pi }{4}} {\frac{{\cancel{\sqrt {\cos 2\theta }} \sin \theta }}{{\cancel{\sqrt {\cos 2\theta }} }}d\theta } = 4\pi {a^2}\int\limits_0^{\frac{\pi }{4}} {\sin \theta d\theta } = 4\pi {a^2}\left. {\left( { - \cos \theta } \right)} \right|_0^{\frac{\pi }{4}} = 4\pi {a^2}\left( { - \frac{{\sqrt 2 }}{2} + 1} \right) = 2\pi {a^2}\left( {2 - \sqrt 2 } \right).$

See more problems on Page 2.