Geometric Applications of Line Integrals
Solved Problems
Example 3.
Find the length of the space curve parameterized by \[\mathbf{r}\left( t \right) = \left( {3t,3{t^2},2{t^3}} \right),\] where \(0 \le t \le 1.\)
Solution.
Using the formula
\[L = \int\limits_C {ds} = \int\limits_\alpha ^\beta {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2} + {{\left( {\frac{{dz}}{{dt}}} \right)}^2}} dt},\]
we have
\[L = \int\limits_C {ds} = \int\limits_0^1 {\sqrt {{3^2} + {{\left( {6t} \right)}^2} + {{\left( {6{t^2}} \right)}^2}} dt} = \int\limits_0^1 {\sqrt {9 + 36{t^2} + 36{t^4}} dt} = \int\limits_0^1 {\sqrt {{{\left( {3 + 6{t^2}} \right)}^2}} dt} = \int\limits_0^1 {\left( {3 + 6{t^2}} \right)dt} = \left. {\left( {3t + 2{t^3}} \right)} \right|_0^1 = 5.\]
Example 4.
Find the arc length of the cycloid parameterized by
\[\mathbf{r}\left( t \right) = \big( {a\left( {t - \sin t} \right),} {a\left( {1 - \cos t} \right)} \big)\]
for \(0 \le t \le 2\pi\) (Figure \(5\text{).}\)
Solution.
Figure 5. We use the formula
\[L = \int\limits_C {ds} = \int\limits_\alpha ^\beta {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} .\]
Here the derivatives are
\[\frac{{dx}}{{dt}} = \frac{{d\left( {a\left( {t - \sin t} \right)} \right)}}{{dt}} = a\left( {1 - \cos t} \right),\]
\[\frac{{dy}}{{dt}} = \frac{{d\left( {a\left( {1 - \cos t} \right)} \right)}}{{dt}} = a\sin t.\]
Then the length of the cycloid is
\[L = \int\limits_0^2 {\sqrt {{{\left( {a\left( {1 - \cos t} \right)} \right)}^2} + {{\left( {a\sin t} \right)}^2}} dt} = \int\limits_0^2 {\sqrt {1 - 2\cos t + {{\cos }^2}t + {{\sin }^2}t} \,dt} = \int\limits_0^2 {\sqrt {2 - 2\cos t} \,dt} = a\sqrt 2 \int\limits_0^2 {\sqrt {1 - \cos t} \,dt} = a\sqrt 2 \int\limits_0^2 {\sqrt {2{{\sin }^2}\left( {\frac{t}{2}} \right)} dt} = 2a\int\limits_0^2 {\sin \frac{t}{2}dt} = 2a\left[ {\left. {\left( { - \frac{{\cos \frac{t}{2}}}{{\frac{1}{2}}}} \right)} \right|_{t = 0}^{2\pi }} \right] = 4a\left( { - \cos \pi + \cos 0} \right) = 8a.\]
Example 5.
Calculate the length of the parabola \[y = {x^2}\] for \(0 \le x \le 1.\)
Solution.
Applying the formula
\[L = \int\limits_a^b {\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx} ,\]
we obtain
\[L = \int\limits_0^1 {\sqrt {1 + {{\left[ {\frac{{d\left( {{x^2}} \right)}}{{dx}}} \right]}^2}} dx} = \int\limits_0^1 {\sqrt {1 + {{\left( {2x} \right)}^2}} dx} = \int\limits_0^1 {\sqrt {1 + 4{x^2}} dx} .\]
To find this integral we make the substitution \(x = {\frac{1}{2}}\tan t,\) \(dx = {\frac{1}{2}} {\frac{{dt}}{{{{\cos }^2}t}}}.\) Hence, \(2x = \tan t,\) \(t = \arctan \left( {2x} \right).\) When \(x = 0\), we have \(t = \arctan 0 = 0,\) and when \(x = 1,\) we get respectively \(t = \arctan 2.\)
Then the length of the parabola is
\[L = \int\limits_0^{\arctan 2} {\sqrt {1 + {{\tan }^2}t} \cdot \frac{{dt}}{{2{{\cos }^2}t}}} - \frac{1}{2}\int\limits_0^{\arctan 2} {\frac{{dt}}{{{{\cos }^3}t}}} = \frac{1}{2}\int\limits_0^{\arctan 2} {\frac{{\cos tdt}}{{{{\cos }^4}t}}} = \frac{1}{2}\int\limits_0^{\arctan 2} {\frac{{d\left( {\sin t} \right)}}{{{{\left( {1 - {\sin^2}t} \right)}^2}}}} .\]
Make one more change of variable. Let now \(\sin t = z.\) If \(t = 0,\) then we have \(z = 0.\) If \(t = \arctan 2,\) we get
\[z = \sin \left( {\arctan 2} \right) = \sqrt {\frac{{{{\left[ {\tan \left( {\arctan 2} \right)} \right]}^2}}}{{{{\left[ {\tan \left( {\arctan 2} \right)} \right]}^2} + 1}}} = \sqrt {\frac{{{2^2}}}{{{2^2} + 1}}} = \frac{2}{{\sqrt 5 }}.\]
We used here the trigonometric identity
\[\sin \alpha = \sqrt {\frac{{{{\tan }^2}\alpha }}{{{{\tan }^2}\alpha + 1}}} .\]
As a result, the length of the curve is
\[L = \frac{1}{2}\int\limits_0^{\frac{2}{{\sqrt 5 }}} {\frac{{dz}}{{{{\left( {1 - {z^2}} \right)}^2}}}} = \frac{1}{2}\int\limits_0^{\frac{2}{{\sqrt 5 }}} {\frac{{dz}}{{{{\left( {1 - z} \right)}^2}{{\left( {1 + z} \right)}^2}}}} .\]
Decompose the integrand into partial functions.
\[
\frac{1}{{{{\left( {1 - z} \right)}^2}{{\left( {1 + z} \right)}^2}}}
= \frac{A}{{{{\left( {1 - z} \right)}^2}}} + \frac{B}{{1 - z}}
+ \frac{C}{{{{\left( {1 + z} \right)}^2}}} + \frac{D}{{1 + z}},\]
\[
1 = A{\left( {1 + z} \right)}^2 + B{\left( {1 - z} \right)}{\left( {1 + z} \right)}^2
+ {C{\left( {1 - z} \right)}^2} + D{\left( {1 + z} \right)}{\left( {1 - z} \right)}^2,\]
\[
1 = A\left( {1 + 2z + {z^2}} \right) + \left( {B - Bz} \right)\left( {1 + 2z + {z^2}} \right)
+ C\left( {1 - 2z + {z^2}} \right) + \left( {D + Dz} \right)\left( {1 - 2z + {z^2}} \right),\]
\[
1 = A + 2Az + A{z^2}
+ B - Bz + 2Bz - 2B{z^2} + B{z^2} - B{z^3}
+ C - 2Cz + C{z^2}
+ D + Dz - 2Dz - 2D{z^2} + D{z^2} + D{z^3}.\]
Consequently,
\[\left\{ \begin{array}{l}
A + B + C + D = 1\\
2A + B - 2C - 2D = 0\\
A - B + C - D = 0\\
- B + D = 0
\end{array} \right..\]
Solving this system of equations we find the coefficients:
\[A = B = C = D = \frac{1}{4}.\]
Thus,
\[L = \frac{1}{2}\int\limits_0^{\frac{2}{{\sqrt 5 }}} {\frac{{dz}}{{{{\left( {1 - z} \right)}^2}{{\left( {1 + z} \right)}^2}}}} = \frac{1}{8}\left[ {\int\limits_0^{\frac{2}{{\sqrt 5 }}} {\frac{{dz}}{{{{\left( {1 - z} \right)}^2}}}} + \int\limits_0^{\frac{2}{{\sqrt 5 }}} {\frac{{dz}}{{1 - z}}} + \int\limits_0^{\frac{2}{{\sqrt 5 }}} {\frac{{dz}}{{{{\left( {1 + z} \right)}^2}}}} + \int\limits_0^{\frac{2}{{\sqrt 5 }}} {\frac{{dz}}{{1 + z}}} } \right] = \frac{1}{8}\left[ {\left. {\left( {\frac{1}{{1 - z}} - \ln \left| {1 - z} \right| - \frac{1}{{1 + z}} + \ln \left| {1 + z} \right|} \right)} \right|_0^{\frac{2}{{\sqrt 5 }}}} \right] = \frac{1}{8}\left[ {\left. {\left( {\frac{{2z}}{{1 - {z^2}}} - \ln \left| {\frac{{1 + z}}{{1 - z}}} \right|} \right)} \right|_0^{\frac{2}{{\sqrt 5 }}}} \right] = \frac{1}{8}\left[ {\frac{{2 \cdot \frac{2}{{\sqrt 5 }}}}{{1 - {{\left( {\frac{2}{{\sqrt 5 }}} \right)}^2}}} - \ln \left| {\frac{{1 + \frac{2}{{\sqrt 5 }}}}{{1 - \frac{2}{{\sqrt 5 }}}}} \right|}\right] = \frac{1}{8}\left( {4\sqrt 5 + \ln \frac{{\sqrt 5 + 2}}{{\sqrt 5 - 2}}} \right) = \frac{{\sqrt 5 }}{2} + \frac{1}{8}\ln \frac{{{{\left( {\sqrt 5 + 2} \right)}^2}}}{{5 - 4}} = \frac{{\sqrt 5 }}{2} + \frac{1}{4}\ln \left( {\sqrt 5 + 2} \right) \approx 1,48.\]
Example 6.
Find the length of the cardioid given in polar coordinates by the equation \[r = 5\left( {1 + \cos \theta } \right)\] (Figure \(6\)).
Solution.
Figure 6. We use the formula
\[L = \int\limits_\alpha ^\beta {\sqrt {{r^2} + {{\left( {\frac{{dr}}{{d\theta }}} \right)}^2}} d\theta } .\]
The length of the cardioid is expressed in the form
\[L = \int\limits_0^{2\pi } {\left[ {{{\left( {5\left( {1 + \cos \theta } \right)} \right)}^2} + {{\left( {\frac{{d\left( {5\left( {1 + \cos \theta } \right)} \right)}}{{d\theta }}} \right)}^2}}\right]^{\frac{1}{2}} d\theta }
= 5\sqrt 2 \int\limits_0^{2\pi } {\sqrt {1 + \cos \theta } d\theta }
= 5\sqrt 2 \int\limits_0^{2\pi } {\sqrt {2{{\left( {\cos\frac{\theta }{2}} \right)}^2}} d\theta }
= 10\int\limits_0^{2\pi } {\sqrt {{{\left( {\cos\frac{\theta }{2}} \right)}^2}} d\theta } .\]
We notice here that \({\cos {\frac{\theta }{2}}} \ge 0\) when \(0 \le \theta \le \pi,\) and \({\cos {\frac{\theta }{2}}} \le 0\) when \(\pi \le \theta \le 2\pi.\) Hence,
\[\sqrt {{{\left( {\cos\frac{\theta }{2}} \right)}^2}} = \cos\frac{\theta }{2}\;\; \text{for}\;\;0 \le \theta \le \pi ,\]
\[\sqrt {{{\left( {\cos\frac{\theta }{2}} \right)}^2}} = -\cos\frac{\theta }{2}\;\; \text{for}\;\;\pi \le \theta \le 2\pi .\]
Writing the latter integral as the sum of two integrals, we find the length of cardioid:
\[L = 10\int\limits_0^{2\pi } {\sqrt {{{\left( {\cos\frac{\theta }{2}} \right)}^2}} d\theta }
= 10\left[ {\int\limits_0^\pi {\cos\frac{\theta }{2}d\theta } + \int\limits_\pi ^{2\pi } {\left( { - \cos\frac{\theta }{2}} \right)d\theta } } \right]
= 10\left[ {\left. {\left( {\frac{{\sin\frac{\theta }{2}}}{{\frac{1}{2}}}} \right)} \right|_{\theta = 0}^\pi - \left. {\left( {\frac{{\sin\frac{\theta }{2}}}{{\frac{1}{2}}}} \right)} \right|_{\theta = \pi }^{2\pi }} \right]
= 20\left[ {\left( {\sin \frac{\pi }{2} - \sin 0} \right) - \left( {\sin \pi - \sin \frac{\pi }{2}} \right)} \right]
= 20\left[ {\left( {1 - 0} \right) - \left( {0 - 1} \right)} \right] = 40.\]
Example 7.
Find the area of the region bounded by the hyperbola \(y = {\frac{1}{x}},\) the \(x\)-axis, and the vertical lines \(x = 1,\) \(x = 2\) (Figure \(7\)).
Solution.
Figure 7. We calculate the area using the line integral:
\[S = - \oint\limits_C {ydx} = - \int\limits_{AB} {ydx} - \int\limits_{BD} {ydx} - \int\limits_{DE} {ydx} - \int\limits_{EA} {ydx} .\]
Calculate each of the integrals separately.
\[- \int\limits_{AB} {ydx} = - \int\limits_1^2 {0 \cdot dx} = 0,\]
\[- \int\limits_{BD} {ydx} = - \int\limits_0^0 {ydx} = 0,\]
\[- \int\limits_{DE} {ydx} = - \int\limits_2^1 {\frac{{dx}}{x}} = \left. {\left( { - \ln x} \right)} \right|_2^1 = - \ln 1 + \ln 2 = \ln 2,\]
\[- \int\limits_{EA} {ydx} = - \int\limits_0^0 {ydx} = 0.\]
Hence, the area of the region is
\[S = \ln 2.\]
Example 8.
Find the area of the region bounded by the ellipse
\[x = a\cos t, y = b\sin t, 0 \le t \le 2\pi \]
(Figure \(8\)).
Solution.
Figure 8. First we apply the formula
\[S = \oint\limits_C {xdy} = \int\limits_\alpha ^\beta {x\left( t \right){\frac{{dy}}{{dt}}} dt} .\]
This yields
\[S = \int\limits_0^{2\pi } {a\cos t\frac{{d\left( {b\sin t} \right)}}{{dt}}dt} = ab\int\limits_0^{2\pi } {{{\cos }^2}tdt} = ab\int\limits_0^{2\pi } {\frac{{1 + \cos 2t}}{2}dt} = \frac{{ab}}{2}\left[ {\left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_0^{2\pi }} \right] = \frac{{ab}}{2}\left[ {\left( {2\pi + 0} \right) - 0} \right] = \pi ab.\]
We can also get the result using the other two formulas:
\[S = - \oint\limits_C {ydx} = - \int\limits_0^{2\pi } {y\left( t \right)\frac{{dx}}{{dt}}dt} = - \int\limits_0^{2\pi } {b\sin t\frac{{d\left( {a\cos t} \right)}}{{dt}}dt} = - ab\int\limits_0^{2\pi } {\left( { - {{\sin }^2}t} \right)dt} = ab\int\limits_0^{2\pi } {\frac{{1 - \cos 2t}}{2}dt} = \frac{{ab}}{2}\left[ {\left. {\left( {t - \frac{{\sin 2t}}{2}} \right)} \right|_0^{2\pi }} \right] = \frac{{ab}}{2}\left[ {\left( {2\pi - 0} \right) - 0} \right] = \pi ab.\]
\[S = \frac{1}{2}\oint\limits_C {xdy - ydx} = \frac{1}{2}\int\limits_0^{2\pi } {\left[ {x\left( t \right)\frac{{dy}}{{dt}} - y\left( t \right)\frac{{dx}}{{dt}}} \right]dt} = \frac{1}{2}\int\limits_0^{2\pi } {\left[ {a\cos t\frac{{d\left( {b\sin t} \right)}}{{dt}} - b\sin t\frac{{d\left( {a\cos t} \right)}}{{dt}}} \right]dt} = \frac{{ab}}{2}\int\limits_0^{2\pi } {\left( {{{\cos }^2}t + {\sin^2}t} \right)dt} = \frac{{ab}}{2}\int\limits_0^{2\pi } {dt} = \frac{{ab}}{2} \cdot 2\pi = \pi ab.\]
Example 9.
Find the volume of the solid formed by rotating the region \(R\) bounded by the curve \(y = 2 - \sin x\) and the lines \(x = 0,\) \(x = 2\pi,\) \(y = 0\) about the \(x\)-axis.
Solution.
The given region \(R\) is shown in Figure \(9.\)
Figure 9. The volume of the solid is
\[V = - \pi \oint\limits_C {{y^2}dx} = - \pi \left[ {\int\limits_{OA} {{y^2}dx} + \int\limits_{AB} {{y^2}dx} + \int\limits_{BD} {{y^2}dx} + \int\limits_{DO} {{y^2}dx} } \right].\]
Calculate the line integrals:
\[\int\limits_{OA} {{y^2}dx} = \int\limits_0^{2\pi } {0dx} = 0,\]
\[\int\limits_{AB} {{y^2}dx} = \int\limits_{2\pi }^{2\pi } {{y^2}dx} = 0,\]
\[\int\limits_{BD} {{y^2}dx} = \int\limits_{2\pi }^0 {{{\left( {2 - \sin x} \right)}^2}dx} = - \int\limits_0^{2\pi } {{{\left( {2 - \sin x} \right)}^2}dx} = - \int\limits_0^{2\pi } {\left( {4 - 4\sin x + {{\sin }^2}x} \right)dx} = - \int\limits_0^{2\pi } {\left( {4 - 4\sin x + \frac{{1 - \cos 2x}}{2}} \right)dx} = \int\limits_0^{2\pi } {\left( {4\sin x - \frac{9}{2} + \frac{{\cos 2x}}{2}} \right)dx} = \left. {\left( { - 4\cos x - \frac{9}{2}x + \frac{{\sin 2x}}{2}} \right)} \right|_0^{2\pi } = \left( { - 4 - 9\pi + 0} \right) - \left( { - 4 - 0 + 0} \right) = - 9\pi ,\]
\[\int\limits_{DO} {{y^2}dx} = \int\limits_0^0 {{y^2}dx} = 0.\]
Hence, the volume of the solid is
\[V = - \pi \left( {0 + 0 - 9\pi + 0} \right) = 9{\pi ^2}.\]
Example 10.
Find the volume of the ellipsoid formed by revolving ellipse with semi-axis \(a\) and \(b\) about the \(x\)-axis (Figure \(10\text{).}\)
Solution.
Figure 10. The parametric equations of the ellipse are
\[x = a\cos t,\;\; y = b\sin t.\]
We can consider the upper half of the ellipse for \(y \ge 0.\) Then the volume of the ellipsoid with the semi-axis \(a, b, b\) is given by
\[V = - \pi \int\limits_C {{y^2}dx}
= - \pi \int\limits_{AOB} {{y^2}dx} - \pi \int\limits_{BOA} {{y^2}dx}
= - \pi \int\limits_{ - a}^a {{0^2}dx} - \pi \int\limits_a^{ - a} {{y^2}dx} = - \pi \int\limits_a^{ - a} {{y^2}dx} ,\]
where \(y\left( x \right)\) implies the equation of the upper half of the ellipse. In parametric form we obtain
\[V = - \pi \int\limits_0^\pi {{{\left( {y\left( t \right)} \right)}^2}\frac{{dx}}{{dt}}dt} = - \pi \int\limits_0^\pi {{{\left( {b\sin t} \right)}^2}\frac{{d\left( {a\cos t} \right)}}{{dt}}dt} = - \pi a{b^2}\int\limits_0^\pi {{{\sin }^2}t\left( { - \sin t} \right)dt} = - \pi a{b^2}\int\limits_0^\pi {{{\sin }^2}t\,d\left( {\cos t} \right)} = \pi a{b^2}\int\limits_0^\pi {\left( {{\cos^2}t - 1} \right)d\left( {\cos t} \right)} = \pi a{b^2}\left[ {\left. {\left( {\frac{{{{\cos }^3}t}}{3} - \cos t} \right)} \right|_0^\pi } \right] = \pi a{b^2}\left[ {\left( {\frac{{{{\cos }^3}\pi }}{3} - \cos \pi } \right) - \left( {\frac{{{{\cos }^3}0}}{3} - \cos 0} \right)} \right] = \pi a{b^2}\left[ {\left( { - \frac{1}{3} + 1} \right) - \left( {\frac{1}{3} - 1} \right)} \right] = \frac{{4\pi a{b^2}}}{3}.\]
In particular, the volume of a sphere when \(a = b = R\) is
\[V = {\frac{{4\pi {R^3}}}{3}}.\]
