Calculus

Applications of Integrals

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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 (x), y (t), and y (θ) are smooth non-negative functions on the given interval.

  1. If the curve y = f (x), a ≤ x ≤ b is rotated about the x-axis, then the surface area is given by
    \[A = 2\pi \int\limits_a^b {f\left( x \right)\sqrt {1 + {{\left[ {f^\prime\left( x \right)} \right]}^2}} dx} .\]
    Surface of revolution obtained by rotating the curve y=f(x) on the interval [a,b] around the x-axis.
    Figure 1.
  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
    \[A = 2\pi \int\limits_c^d {y\sqrt {1 + {{\left[ {g^\prime\left( y \right)} \right]}^2}} dy} .\]
    Surface of revolution obtained by rotating the curve x=g(y) on the interval [c,d] around the x-axis.
    Figure 2.
  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)
    \[A = 2\pi \int\limits_\alpha ^\beta {y\left( t \right)\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .\]
    Surface obtained by rotating the parametric curve x=x(t), y=y(t) around the x-axis.
    Figure 3.
  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)
    \[A = 2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\sin \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .\]
    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
    \[A = 2\pi \int\limits_a^b {x\sqrt {1 + {{\left[ {f^\prime\left( x \right)} \right]}^2}} dx} .\]
    Surface obtained by rotating the curve y=f(x) around the y-axis.
    Figure 5.
  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
    \[A = 2\pi \int\limits_c^d {g\left( y \right)\sqrt {1 + {{\left[ {g^\prime\left( y \right)} \right]}^2}} dy} .\]
    Surface obtained by rotating the curve x=g(y) around the y-axis.
    Figure 6.
  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)
    \[A = 2\pi \int\limits_\alpha ^\beta {x\left( t \right)\sqrt {{{\left[ {x^\prime\left( t \right)} \right]}^2} + {{\left[ {y^\prime\left( t \right)} \right]}^2}} dt} .\]
    Surface obtained by rotating the parametric curve x=x(t), y=y(t) about the y-axis.
    Figure 7.
  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)
    \[A = 2\pi \int\limits_\alpha ^\beta {r\left( \theta \right)\cos \theta \sqrt {{{\left[ {r\left( \theta \right)} \right]}^2} + {{\left[ {r^\prime\left( \theta \right)} \right]}^2}} d\theta } .\]
    Surface obtained by rotating the polar curve r=r(theta) about the y-axis.
    Figure 8.

Solved Problems

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

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

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.

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 \(f\left( x \right) = a\cosh \frac{x}{a},\) 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.

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:

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

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

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

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.

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