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)

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

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

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



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)

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

Solved Problems

Click or tap a problem to see the solution.

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,

\[A = \pi R\ell.\]

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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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