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

3. 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
4. $A = 2\pi \int\limits_c^d {y\sqrt {1 + {{\left[ {g^\prime\left( y \right)} \right]}^2}} dy} .$

5. 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)
6. $A = 2\pi \int\limits_\alpha ^\beta {y\left( t \right)\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .$

7. 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)
8. ${A \text{ = }}\kern0pt{2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\sin \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .}$

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

3. 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
4. $A = 2\pi \int\limits_c^d {g\left( y \right)\sqrt {1 + {{\left[ {g^\prime\left( y \right)} \right]}^2}} dy} .$

5. 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)
6. $A = 2\pi \int\limits_\alpha ^\beta {x\left( t \right)\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .$

7. 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)
8. ${A \text{ = }}\kern0pt{2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\cos \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } }$

## 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 $$y = a\cosh \large{\frac{x}{a}}\normalsize$$ 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 5

The cardioid $$r = 1 + \cos \theta$$ rotates around the polar axis. Find the area of the resulting surface.

### Example 6

The infinite curve $$y = {e^{ – x}},$$ where $$x \ge 0,$$ is rotated around the $$x-$$axis. Find the area of the resulting surface.

### Example 7

Find the area of the surface formed by rotating the parabola $$y = 1 – {x^2}$$ on the interval $$\left[ {0,1} \right]$$ around the $$y-$$axis.

### Example 8

Find the area of the surface obtained by rotating the curve $$y = \sqrt[3]{x}$$ on the interval $$\left[ {0,1} \right]$$ around the $$y-$$axis.

### Example 9

Find the area of the surface obtained by rotating the circle $$r = 2\sin \theta$$ about the $$y-$$axis.

### Example 10

One arch of the cycloid $$x = \theta – \sin \theta,$$ $$y = 1 – \cos \theta$$ is rotated around the $$y-$$axis. Calculate 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 $$a = 0,$$ $$b = H,$$ $$f\left( x \right) = kx = \large{\frac{R}{H}}\normalsize x,$$ $$f^\prime\left( x \right) = k = \large{\frac{R}{H}}\normalsize,$$ we obtain

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

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

$A = \pi R\ell.$

### Example 2.

The catenary line $$y = a\cosh \large{\frac{x}{a}}\normalsize$$ 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 $$f\left( x \right) = a\cosh \large{\frac{x}{a}}\normalsize,$$ 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:

${1 + {\left( {\sinh x} \right)^2} = {\left( {\cosh x} \right)^2},\;\;}\kern0pt{{\left( {\cosh x} \right)^2} = \frac{1}{2}\left[ {\cosh \left( {2x} \right) + 1} \right].}$

This yields:

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

### 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:

${\left( {{{\cos }^3}t} \right)’ = – 3\,{\cos ^2}t\sin t,\;\;}\kern0pt{\left( {{{\sin }^3}t} \right)’ = 3\,{\sin ^2}t\cos t,}$

and simplify the radicand:

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

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

${A \text{ = }}\kern0pt{2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\sin \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } }$

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

${A \text { = }}\kern0pt{4\pi \int\limits_0^{\frac{\pi }{4}} {r\left( \theta \right)\sin \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .}$

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

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

Let’s 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

${A \text { = }}\kern0pt{4\pi \int\limits_0^{\frac{\pi }{4}} {\left( {a\sqrt {\cos 2\theta } \sin \theta \cdot \frac{a}{{\sqrt {\cos 2\theta } }}} \right)d\theta } }={ 4\pi {a^2}\int\limits_0^{\frac{\pi }{4}} {\frac{{\cancel{\sqrt {\cos 2\theta }} \sin \theta }}{{\cancel{\sqrt {\cos 2\theta }} }}d\theta } }={ 4\pi {a^2}\int\limits_0^{\frac{\pi }{4}} {\sin \theta d\theta } }={ 4\pi {a^2}\left. {\left( { – \cos \theta } \right)} \right|_0^{\frac{\pi }{4}} }={ 4\pi {a^2}\left( { – \frac{{\sqrt 2 }}{2} + 1} \right) }={ 2\pi {a^2}\left( {2 – \sqrt 2 } \right).}$

### Example 5.

The cardioid $$r = 1 + \cos \theta$$ rotates around the polar axis. Find the area of the resulting surface.

Solution.

As the curve is defined in polar coordinates and rotated about the $$x-$$axis, we calculate the surface area by the formula

${A \text{ = }}\kern0pt{2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\sin \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } }$

Here

${r\left( \theta \right) = 1 + \cos \theta ,\;\;}\kern0pt{r^\prime\left( \theta \right) = \left( {1 + \cos \theta } \right)^\prime }={ – \sin \theta .}$

Simplify the expression under the square root sign:

${{\left[ {r\left( \theta \right)} \right]^2} + {\left[ {r’\left( \theta \right)} \right]^2} }={ {\left( {1 + \cos \theta } \right)^2} + {\left( { – \sin \theta } \right)^2} }={ 1 + 2\cos \theta + {\cos ^2}\theta + {\sin ^2}\theta }={ 2\left( {1 + \cos \theta } \right).}$

Let’s recall now the double angle identities:

${1 + \cos \theta = 2{\cos ^2}\frac{\theta }{2},\;\;}\kern0pt{\sin \theta = 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}.}$

Substituting these formulas we can write the integral in the form

${A \text { = }}\kern0pt{2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\sin \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } }={ 2\pi \int\limits_0^\pi {\left( {1 + \cos \theta } \right)\sin \theta \sqrt {2\left( {1 + \cos \theta } \right)} d\theta } }={ 2\pi \int\limits_0^\pi {\left( {2{{\cos }^2}\frac{\theta }{2} \cdot 2\sin \frac{\theta }{2}\cos \frac{\theta }{2} \cdot }\right.}}\kern0pt{{\left.{2\cos \frac{\theta }{2}} \right)d\theta } }={ 16\pi \int\limits_0^\pi {{{\cos }^4}\frac{\theta }{2}\sin \frac{\theta }{2}d\theta .} }$

It’s convenient to change variable:

${\cos \frac{\theta }{2} = z,}\;\; \Rightarrow {- \frac{1}{2}\sin \frac{\theta }{2}d\theta = dz,}\;\; \Rightarrow {\sin \frac{\theta }{2}d\theta = – 2dz.}$

When $$\theta = 0,$$ $$z = 1$$, and when $$\theta = \pi,$$ $$z = 0.$$ Hence, the surface area is equal to

${A = 16\pi \int\limits_0^\pi {{{\cos }^4}\frac{\theta }{2}\sin \frac{\theta }{2}d\theta } }={ 16\pi \int\limits_1^0 {{z^4}\left( { – 2dz} \right)} }={ 32\pi \int\limits_0^1 {{z^4}dz} }={ 32\pi \cdot \left. {\frac{{{z^5}}}{5}} \right|_0^1 }={ \frac{{32\pi }}{5}}$

### Example 6.

The infinite curve $$y = {e^{ – x}},$$ where $$x \ge 0,$$ is rotated around the $$x-$$axis. Find the area of the resulting surface.

Solution.

The surface area is determined by the formula

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

Since $$f^\prime\left( x \right) = \left( {{e^{ – x}}} \right)^\prime = – {e^{ – x}},$$ we can write

${A = 2\pi \int\limits_0^\infty {{e^{ – x}}\sqrt {1 + {{\left[ { – {e^{ – x}}} \right]}^2}} dx} }={ 2\pi \int\limits_0^\infty {{e^{ – x}}\sqrt {1 + {e^{ – 2x}}} dx} .}$

To make the integration simpler, we change the variable:

${z = {e^{ – x}},}\;\; \Rightarrow {dz = – {e^{ – x}}dx,}\;\; \Rightarrow {{e^{ – x}}dx = – dz.}$

When $$x = 0,$$ $$z = 1,$$ and when $$x = \infty,$$ $$z = 0.$$ So we have

${A = 2\pi \int\limits_1^0 {\left( { – \sqrt {1 + {z^2}} } \right)dz} }={ 2\pi \int\limits_0^1 {\sqrt {1 + {z^2}} dz} .}$

We can evaluate this integral using the trigonometric substitution $$z = \tan t.$$

${z = \tan t,\;\;} \Rightarrow {\sqrt {1 + {z^2}} = \sqrt {1 + {{\tan }^2}t} = \sec t,\;\;}\kern0pt{dz = {\sec ^2}t dt.}$

When $$z = 0,$$ $$t = 0,$$ and when $$z = 1,$$ $$t = \large{\frac{\pi }{4}}\normalsize.$$ Hence

${A = 2\pi \int\limits_0^1 {\sqrt {1 + {z^2}} dz} }={ 2\pi \int\limits_0^{\frac{\pi }{4}} {\sec t{{\sec }^2}tdt} }={ 2\pi \int\limits_0^{\frac{\pi }{4}} {{{\sec }^3}tdt} .}$

The last integral can be found using the reduction formula

${\int {{{\sec }^3}tdt} }={ \frac{{\sec t\tan t}}{2} + \frac{1}{2}\int {\sec tdt} ,}$

where $$\int {\sec tdt}$$ is a table integral which is equal to

$\int {\sec tdt} = \ln \left| {\tan \left( {\frac{t}{2} + \frac{\pi }{4}} \right)} \right|.$

Thus,

${A = 2\pi \int\limits_0^{\frac{\pi }{4}} {{{\sec }^3}tdt} }={ 2\pi \left. {\left[ {\frac{{\sec t\tan t}}{2} }\right.}+{\left.{ \frac{1}{2}\ln \left| {\tan \left( {\frac{t}{2} + \frac{\pi }{4}} \right)} \right|} \right]} \right|_0^{\frac{\pi }{4}} }={ 2\pi \left[ {\left( {\frac{{\sqrt 2 \cdot 1}}{2} + \frac{1}{2}\ln \tan \frac{{3\pi }}{8}} \right) }\right.}-{\left.{ \left( {0 + \frac{1}{2}\underbrace {\ln \tan \frac{\pi }{4}}_{ = 0}} \right)} \right] }={ \pi \left[ {\sqrt 2 + \ln \tan \frac{{3\pi }}{8}} \right].}$

Given that

${\tan \frac{{3\pi }}{8} = \frac{{\sin \frac{{3\pi }}{4}}}{{1 + \cos \frac{{3\pi }}{4}}} }={ \frac{{\frac{{\sqrt 2 }}{2}}}{{1 – \frac{{\sqrt 2 }}{2}}} }={ \frac{1}{{\sqrt 2 – 1}} }={ \frac{{\sqrt 2 + 1}}{{\left( {\sqrt 2 – 1} \right)\left( {\sqrt 2 + 1} \right)}} }={ 1 + \sqrt 2 ,}$

we get the final answer in the form

${A = \pi \left[ {\sqrt 2 + \ln \tan \frac{{3\pi }}{8}} \right] }={ \pi \left[ {\sqrt 2 + \ln \left( {1 + \sqrt 2 } \right)} \right].}$

### Example 7.

Find the area of the surface formed by rotating the parabola $$y = 1 – {x^2}$$ on the interval $$\left[ {0,1} \right]$$ around the $$y-$$axis.

Solution.

The surface area is determined by the integral

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

Here $$a =0,$$ $$b = 1,$$ $$f^\prime\left( x \right) = \left( {1 – {x^2}} \right)’ = – 2x.$$ Hence

${A = 2\pi \int\limits_0^1 {x\sqrt {1 + {{\left( { – 2x} \right)}^2}} dx} }={ 2\pi \int\limits_0^1 {x\sqrt {1 + 4{x^2}} dx} .}$

We make the substitution:

${1 + 4{x^2} = {t^2},}\;\; \Rightarrow {8xdx = 2tdt,}\;\; \Rightarrow {xdx = \frac{1}{4}tdt.}$

When $$x = 0,$$ $$t = 1,$$ and when $$x = 1,$$ $$t = \sqrt{5}.$$ This yields

${A = 2\pi \int\limits_1^{\sqrt 5 } {\left( {t \cdot \frac{1}{4}t} \right)dt} }={ \frac{\pi }{2}\int\limits_1^{\sqrt 5 } {{t^2}dt} }={ \frac{\pi }{2} \cdot \left. {\frac{{{t^3}}}{3}} \right|_1^{\sqrt 5 } }={ \frac{\pi }{6}\left( {5\sqrt 5 – 1} \right).}$

### Example 8.

Find the area of the surface obtained by rotating the curve $$y = \sqrt[3]{x}$$ on the interval $$\left[ {0,1} \right]$$ around the $$y-$$axis.

Solution.

We rewrite the equation of the curve as a function $$x = g\left( y \right):$$

${y = f\left( x \right) = \sqrt[3]{x},}\;\; \Rightarrow {x = g\left( y \right) = {y^3}.}$

Now we apply the integration formula

$A = 2\pi \int\limits_c^d {g\left( y \right)\sqrt {1 + {{\left[ {g^\prime\left( y \right)} \right]}^2}} dy} ,$

where $$c = 0,$$ $$d = 1,$$ $$g^\prime\left( y \right) = \left( {{y^3}} \right)^\prime = 3{y^2}.$$ Hence,

${A = 2\pi \int\limits_0^1 {{y^3}\sqrt {1 + {{\left[ {3{y^2}} \right]}^2}} dy} }={ 2\pi \int\limits_0^1 {{y^3}\sqrt {1 + 9{y^4}} dy} .}$

Make the substitution

${1 + 9{y^4} = {t^2},}\;\; \Rightarrow {36{y^3}dy = 2tdt,}\;\; \Rightarrow {{y^3}dy = \frac{1}{{18}}tdt.}$

When $$y = 0,$$ $$t =1,$$ and when $$y = 1,$$ $$t = \sqrt{10}.$$ So the integral becomes

${A = 2\pi \int\limits_1^{\sqrt {10} } {\left( {t \cdot \frac{1}{{18}}t} \right)dt} }={ \frac{\pi }{9}\int\limits_1^{\sqrt {10} } {{t^2}dt} }={ \frac{\pi }{9} \cdot \left. {\frac{{{t^3}}}{3}} \right|_1^{\sqrt {10} } }={ \frac{\pi }{{27}}\left( {10\sqrt {10} – 1} \right).}$

### Example 9.

Find the area of the surface obtained by rotating the circle $$r = 2\sin \theta$$ about the $$y-$$axis.

Solution.

Since the curve is given by the polar equation, we use the integration formula

${A \text { = }}\kern0pt{ 2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\cos \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .}$

Integrating from $$0$$ to $$\large{\frac{\pi}{2}}\normalsize$$ and substituting $$r\left( \theta \right) = 2\sin \theta,$$ $$r^\prime\left( \theta \right) = 2\cos \theta,$$ we have

${A \text { = }}\kern0pt{2\pi \int\limits_0^{\frac{\pi }{2}} {2\sin \theta \cos \theta } }\kern0pt{\sqrt {{{\left[ {2\sin \theta } \right]}^2} + {{\left[ {2\cos \theta } \right]}^2}} d\theta }={ 4\pi \int\limits_0^{\frac{\pi }{2}} {\sin 2\theta } \underbrace {\sqrt {{{\sin }^2}\theta + {{\cos }^2}\theta } }_{ = 1}\,d\theta }={ 4\pi \int\limits_0^{\frac{\pi }{2}} {\sin 2\theta } d\theta }={ 4\pi \left. {\left( { – \frac{{\cos 2\theta }}{2}} \right)} \right|_0^{\frac{\pi }{2}} }={ 2\pi \left( { – \cos \pi + \cos 0} \right) }={ 4\pi .}$

### Example 10.

One arch of the cycloid $$x = \theta – \sin \theta,$$ $$y = 1 – \cos \theta$$ is rotated around the $$y-$$axis. Calculate the area of the resulting surface.

Solution.

The curve is given in parametric form. Therefore, we use the following integration formula

${A \text { = }}\kern0pt{2\pi \int\limits_\alpha ^\beta {x\left( \theta \right)\sqrt {{{\left[ {x^\prime\left( \theta \right)} \right]}^2} + {{\left[ {y^\prime\left( \theta \right)} \right]}^2}} d\theta } ,}$

where the parameter $$\theta$$ varies from $$0$$ to $$2\pi.$$

Take the derivatives:

${x^\prime\left( \theta \right) = \left( {\theta – \sin \theta } \right)^\prime }={ 1 – \cos \theta ,}$

${y^\prime\left( \theta \right) = \left( {1 – \cos \theta } \right)^\prime }={ \sin \theta ,}$

and simplify the expression under the root square sign:

${{\left[ {x’\left( \theta \right)} \right]^2} + {\left[ {y’\left( \theta \right)} \right]^2} }={ {\left( {1 – \cos \theta } \right)^2} + {\sin ^2}\theta }={ 1 – 2\cos \theta + \underbrace {{{\cos }^2}\theta + {{\sin }^2}\theta }_{ = 1} }={ 2 – 2\cos \theta }={ 4{\sin ^2}\frac{\theta }{2}.}$

Then the surface area is given by

${A \text { = }}\kern0pt{2\pi \int\limits_0^{2\pi } {\left[ {\left( {\theta – \sin \theta } \right) \cdot 2\sin \frac{\theta }{2}} \right]d\theta } }={ 4\pi \left[ {\int\limits_0^{2\pi } {\theta \sin \frac{\theta }{2}d\theta } }\right.}-{\left.{ \int\limits_0^{2\pi } {\sin \theta \sin \frac{\theta }{2}d\theta } } \right] }={ 4\pi \left[ {{I_1} – {I_2}} \right].}$

We calculate the first integral using integration by parts:

${{I_1} = \int\limits_0^{2\pi } {\theta \sin \frac{\theta }{2}d\theta } }={ \left[ {\begin{array}{*{20}{l}} {u = \theta }\\ {dv = \sin \frac{\theta }{2}d\theta }\\ {u’ = 1}\\ {v = – 2\cos \frac{\theta }{2}} \end{array}} \right] }={ – \left. {2\theta \cos \frac{\theta }{2}} \right|_0^{2\pi } }-{ \int\limits_0^{2\pi } {\left( { – 2\cos \frac{\theta }{2}} \right)d\theta } }={ – \left. {2\theta \cos \frac{\theta }{2}} \right|_0^{2\pi } }+{ 2\int\limits_0^{2\pi } {\cos \frac{\theta }{2}d\theta } }={ – \left. {2\theta \cos \frac{\theta }{2}} \right|_0^{2\pi } }+{ \left. {4\sin \frac{\theta }{2}} \right|_0^{2\pi } }={ \left. {\left[ {4\sin \frac{\theta }{2} – 2\theta \cos \frac{\theta }{2}} \right]} \right|_0^{2\pi } }={ 4\pi .}$

Consider now the second integral. Notice that

${\int {\sin \theta \sin \frac{\theta }{2}d\theta } }={ 2\int {{{\sin }^2}\frac{\theta }{2}\cos \frac{\theta }{2}d\theta } }={ 4\int {{{\sin }^2}\frac{\theta }{2}d\left( {\sin \frac{\theta }{2}} \right)} }={ \frac{4}{3}{\sin ^3}\frac{\theta }{2} + C.}$

Hence,

${{I_2} = \int\limits_0^{2\pi } {\sin \theta \sin \frac{\theta }{2}d\theta } }={ \frac{4}{3}\left. {{{\sin }^3}\frac{\theta }{2}} \right|_0^{2\pi } }={ 0.}$

So the area of the surface is

$A = 4\pi \left[ {{I_1} – {I_2}} \right] = 16{\pi ^2}.$