Calculus

Applications of Integrals

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Area of a Surface of Revolution

Solved Problems

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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 ≥ 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 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

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

Here

\[r\left( \theta \right) = 1 + \cos \theta ,\;\; r^\prime\left( \theta \right) = \left( {1 + \cos \theta } \right)^\prime = - \sin \theta .\]

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:

\[1 + \cos \theta = 2\,{\cos ^2}\frac{\theta }{2},\;\; \sin \theta = 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}.\]

Substituting these formulas we can write the integral in the form

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

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 \(z = \tan t.\)

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

When \(z = 0,\) \(t = 0,\) and when \(z = 1,\) \(t = \frac{\pi }{4}.\) 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,

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

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 \(a =0,\) \(b = 1,\) \(f^\prime\left( x \right) = \left( {1 - {x^2}} \right)^\prime = - 2x.\) 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

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

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

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

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

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

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

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

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

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