Calculus

Integration of Functions

Integration of Functions Logo

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

      Surface of revolution obtained by rotating the curve y=f(x) on the interval [a,b] around the x-axis.
      Figure 1.
    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. \[A = 2\pi \int\limits_c^d {y\sqrt {1 + {{\left[ {g^\prime\left( y \right)} \right]}^2}} dy} .\]

      Surface of revolution obtained by rotating the curve x=g(y) on the interval [c,d] around the x-axis.
      Figure 2.
    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. \[A = 2\pi \int\limits_\alpha ^\beta {y\left( t \right)\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .\]

      Surface obtained by rotating the parametric curve x=x(t), y=y(t) around the x-axis.
      Figure 3.
    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. \[{A \text{ = }}\kern0pt{2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\sin \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .}\]

      Surface obtained by rotating the polar curve r=r(theta) about the x-axis.
      Figure 4.

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

      Surface obtained by rotating the curve y=f(x) around the y-axis.
      Figure 5.
    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. \[A = 2\pi \int\limits_c^d {g\left( y \right)\sqrt {1 + {{\left[ {g^\prime\left( y \right)} \right]}^2}} dy} .\]

      Surface obtained by rotating the curve x=g(y) around the y-axis.
      Figure 6.
    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. \[A = 2\pi \int\limits_\alpha ^\beta {x\left( t \right)\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .\]

      Surface obtained by rotating the parametric curve x=x(t), y=y(t) about the y-axis.
      Figure 7.
    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. \[{A \text{ = }}\kern0pt{2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\cos \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } }\]

      Surface obtained by rotating the polar curve r=r(theta) about the y-axis.
      Figure 8.

  • Solved Problems

    Click 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.

    A right circular conic surface obtained by rotating the line y=Rx/H around the x-axis.
    Figure 9.

    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.

    Catenoid formed by revolving the catenary line y=a*cosh(x/a) around the x-axis.
    Figure 10.

    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.

    Surface obtained by rotating the astroid x=(cos(t))^3, y=((sin(t))^3 around the x-axis.
    Figure 11.

    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.

    Surface obtained by rotating the lemniscate of Bernoulli around the x-axis.
    Figure 12.

    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.

    Surface obtained by revolving the cardioid r=1+cos(x) about the x-axis.
    Figure 13.

    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.

    Surface obtained by rotating the infinite curve y=exp(-x) around the x-axis.
    Figure 14.

    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.

    Surface formed by rotating the parabola y=1-x^2 on the interval [0,1] around the y-axis
    Figure 15.

    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.

    Surface formed by rotating the cubic root function y=sqrt[3](x) around the vertical axis.
    Figure 16.

    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.

    The ball surface obtained by rotating the circle r=2sin(theta) around the y-axis.
    Figure 17.

    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.

    Surface formed by rotating one arch of the cycloid x=theta-sin(theta), y=1-cos(theta) around the y-axis.
    Figure 18.

    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

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