## Calculus (3rd Edition)

Length $\simeq 5.52$
First, we sketch the curve to determine the limits of integration. Since $r = {\cos ^2}\theta$, the rectangular coordinates of a point on the curve corresponding to $\theta$ is given by $\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right)$, $\left( {x,y} \right) = {\cos ^2}\theta \left( {\cos \theta ,\sin \theta } \right)$. For the interval $0 \le \theta \le 2\pi$, we evaluate several points in rectangular coordinates corresponding to $\theta = 0,\frac{\pi }{4},\frac{\pi }{2},...,2\pi$ and list them on the following table: $\begin{array}{*{20}{c}} \theta &{\left( {x,y} \right)}\\ 0&{\left( {1,0} \right)}\\ {\pi /4}&{\left( {0.354,0.354} \right)}\\ {\pi /2}&{\left( {0,0} \right)}\\ {3\pi /4}&{\left( { - 0.354,0.354} \right)}\\ \pi &{\left( { - 1,0} \right)}\\ {5\pi /4}&{\left( { - 0.354, - 0.354} \right)}\\ {3\pi /2}&{\left( {0,0} \right)}\\ {7\pi /4}&{\left( {0.354, - 0.354} \right)}\\ {2\pi }&{\left( {1,0} \right)} \end{array}$ Then we plot the points in rectangular coordinates and sketch the curve by connecting these points. We have $r = f\left( \theta \right) = {\cos ^2}\theta$, ${\ \ }$ $f'\left( \theta \right) = - 2\cos \theta \sin \theta$ Next, we compute the arc length of the curve for $0 \le \theta \le \frac{\pi }{2}$. $s = \mathop \smallint \limits_0^{\pi /2} \sqrt {{{\cos }^4}\theta + 4{{\cos }^2}\theta {{\sin }^2}\theta } {\rm{d}}\theta$ $s = \mathop \smallint \limits_0^{\pi /2} \sqrt {{{\cos }^2}\theta \left( {{{\cos }^2}\theta + 4{{\sin }^2}\theta } \right)} {\rm{d}}\theta$ $s = \mathop \smallint \limits_0^{\pi /2} \cos \theta \sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta + 3{{\sin }^2}\theta } {\rm{d}}\theta$ $s = \mathop \smallint \limits_0^{\pi /2} \cos \theta \sqrt {1 + 3{{\sin }^2}\theta } {\rm{d}}\theta$ Let $u = \sqrt 3 \sin \theta$. So, $du = \sqrt 3 \cos \theta d\theta$. Substituting these in the integral above gives $s = \frac{1}{{\sqrt 3 }}\mathop \smallint \limits_0^{\sqrt 3 } \sqrt {1 + {u^2}} {\rm{d}}u$ Let $u = \tan v$. So, $du = {\sec ^2}vdv$. The integral becomes $s = \frac{1}{{\sqrt 3 }}\mathop \smallint \limits_0^{{{\tan }^{ - 1}}\sqrt 3 } {\sec ^2}v\sqrt {1 + {{\tan }^2}v} {\rm{d}}v$ Since ${\sec ^2}v - {\tan ^2}v = 1$, we get $s = \frac{1}{{\sqrt 3 }}\mathop \smallint \limits_0^{{{\tan }^{ - 1}}\sqrt 3 } {\sec ^3}v{\rm{d}}v$ From Eq. 14 of Section 8.2 (page 402) we have $\smallint {\sec ^m}x{\rm{d}}x = \frac{{\tan x{{\sec }^{m - 2}}x}}{{m - 1}} + \frac{{m - 2}}{{m - 1}}\smallint {\sec ^{m - 2}}{\rm{d}}x$ So, $s = \frac{1}{{\sqrt 3 }}\left( {\frac{{\tan v\sec v}}{2}|_0^{{{\tan }^{ - 1}}\sqrt 3 } + \frac{1}{2}\mathop \smallint \limits_0^{{{\tan }^{ - 1}}\sqrt 3 } \sec v{\rm{d}}v} \right)$ From Eq. 13 of Section 8.2 (page 402) we have $\smallint \sec x{\rm{d}}x = \ln \left| {\sec x + \tan x} \right| + C$ So, $s = \frac{1}{{\sqrt 3 }}\left( {\frac{{\tan v\sec v}}{2}|_0^{{{\tan }^{ - 1}}\sqrt 3 } + \frac{1}{2}\ln \left| {\sec v + \tan v} \right||_0^{{{\tan }^{ - 1}}\sqrt 3 }} \right)$ $s = 1 + \frac{1}{{2\sqrt 3 }}\ln \left( {2 + \sqrt 3 } \right) \simeq 1.38$ By symmetry, the total length of the curve is four times the arc length of the curve for $0 \le \theta \le \frac{\pi }{2}$. Therefore, length $\simeq 4\cdot1.38 \simeq 5.52$