# Thread: Application of integral calculus. Find a length

1. ## Application of integral calculus. Find a length

Help me please to solve these problems on the application of integral calculus to problems of geometry:
1. Find a length of the curve y = ln(cos(x)) +2, 0 =< x =< pi/6.
2. Find a length of the polar curve r = sqrt(2)*exp{fi}, 0 =< fi =< 2pi.
3. Find a length of the parametric curve x = 3(t - sin(t)), y = 3(1 - cos(t)), 0 =< t =< 2pi.
4. Find a square which limited the polar curve r = a*cos(3*fi), 0 =< fi =< 2pi.

For me these problems have been the most difficult some of them, I began to deal with, but my answers were not correct.
Tell me please, what program would be easier to do the drawings for these tasks.

2. Originally Posted by Dext91
Help me please to solve these problems on the application of integral calculus to problems of geometry:

1. Find a length of the curve y = ln(cos(x)) +2, 0 =< x =< pi/6.

So $\displaystyle 1.{\text{ }}y = \ln \cos x + 2,{\text{ }}0 \leqslant x \leqslant \frac{\pi }{6},{\text{ }}L = ?$

$\displaystyle L = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } = \int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}} {\sqrt {1 + {{\left( {\frac{d}{{dx}}\left( {\ln \cos x + 2} \right)} \right)}^2}} }.$

$\displaystyle {\left( {\frac{d}{{dx}}\left( {\ln \cos x + 2} \right)} \right)^2} = {\left( { - \frac{{\sin x}}{{\cos x}}} \right)^2} = \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}.$

$\displaystyle \sqrt {1 + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} = \sqrt {\frac{1}{{{{\cos }^2}x}}} = \frac{1}{{\cos x}}{\text{ }}\left( {0 \leqslant x \leqslant \frac{\pi }{6}} \right).$

$\displaystyle L = \int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}} {\frac{{dx}}{{\cos x}}} = \int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}} {\frac{{\cos x}}{{{{\cos }^2}x}}dx} = - \int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}} {\frac{{\cos x}}{{{{\sin }^2}x - 1}}dx} =$

$\displaystyle = - \int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}} {\frac{{d\left( {\sin x} \right)}}{{{{\sin }^2}x - 1}}} = \left. { - \frac{1}{2}\ln \left| {\frac{{\sin x - 1}}{{\sin x + 1}}} \right|} \right|_0^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}} =$

$\displaystyle = - \frac{1}{2}\left( {\ln \left| {\frac{{{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2} - 1}}{{{1 \mathord{\left/{\vphantom {1 2}} \right.\kern-\nulldelimiterspace} 2} + 1}}} \right| - \ln \left| { - 1} \right|} \right) = - \frac{1}{2}\ln \frac{1}{3} = \frac{{\ln 3}}{2}.$

Look this picture

3. Originally Posted by Dext91
Help me please to solve these problems on the application of integral calculus to problems of geometry:

2. Find a length of the polar curve r = sqrt(2)*exp{fi}, 0 =< fi =< 2pi.

$\displaystyle r = \sqrt 2 {e^\varphi },{\text{ }}0 \leqslant \varphi \leqslant 2\pi ,{\text{ }}L = ?$

$\displaystyle L = \int\limits_\alpha ^\beta {\sqrt {{r^2} + {{r'}^2}} d\varphi } = \int\limits_0^{2\pi } {\sqrt {{{\left( {\sqrt 2 {e^\varphi }} \right)}^2} + {{\left( {\sqrt 2 {e^\varphi }} \right)}^\prime }^2} d\varphi } =$

$\displaystyle = \int\limits_0^{2\pi } {\sqrt {2{e^{2\varphi }} + 2{e^{2\varphi }}} d\varphi } = 2\int\limits_0^{2\pi } {{e^\varphi }d\varphi } = \left. {2{e^\varphi }} \right|_0^{2\pi } = 2{e^{2\pi }} - 2.$

Look this picture

4. Originally Posted by Dext91
Help me please to solve these problems on the application of integral calculus to problems of geometry:

4. Find a square which limited the polar curve r = a*cos(3*fi), 0 =< fi =< 2pi.
So we have $\displaystyle r = a\cos 3\varphi ,{\text{ }}0 \leqslant \varphi \leqslant 2\pi ,{\text{ }}S - ?$

Solution

$\displaystyle \frac{S}{6} = \frac{1}{2}\int\limits_\alpha ^\beta {{r^2}d\varphi } = \frac{{{a^2}}}{2}\int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}} {{{\cos }^2}3\varphi d\varphi } =$

$\displaystyle = \frac{{{a^2}}}{4}\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}} {\left( {1 + \cos 6\varphi } \right)d\varphi } = \left. {\frac{{{a^2}}}{4}\left( {\varphi + \frac{1}{6}\sin 6\varphi } \right)} \right|_0^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}}$$\displaystyle = \frac{{{a^2}}}{4} \cdot \frac{\pi }{6} = \frac{{\pi {a^2}}}{{24}}.$

So $\displaystyle \frac{S}{6} = \frac{{\pi {a^2}}}{{24}} \Leftrightarrow S = \frac{{\pi {a^2}}}{4}.$

Look this picture

5. Originally Posted by Dext91
Help me please to solve these problems on the application of integral calculus to problems of geometry:
3. Find a length of the parametric curve x = 3(t - sin(t)), y = 3(1 - cos(t)), 0 =< t =< 2pi.
$\displaystyle x = 3\left( {t - \sin t} \right),{\text{ }}y = 3\left( {1 - \cos t} \right),{\text{ }}0 \leqslant t \leqslant 2\pi ,{\text{ }}L = ?$

$\displaystyle L = \int\limits_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} = \int\limits_0^{2\pi } {\sqrt {{{\left( {\frac{d}{{dt}}3\left( {t - \sin t} \right)} \right)}^2} + {{\left( {\frac{d}{{dt}}3\left( {1 - \cos t} \right)} \right)}^2}} dt} .$

$\displaystyle {\left( {\frac{d}{{dt}}3\left( {t - \sin t} \right)} \right)^2} = 9{\left( {1 - \cos t} \right)^2}$ and $\displaystyle {\left( {\frac{d}{{dt}}3\left( {1 - \cos t} \right)} \right)^2} = 9{\sin ^2}t.$

$\displaystyle \sqrt {9{{\left( {1 - \cos t} \right)}^2} + 9{{\sin }^2}t} = 3\sqrt {1 - 2\cos t + {{\cos }^2}t + {{\sin }^2}t} = 3\sqrt {2 - 2\cos t} =$

$\displaystyle = 3\sqrt {2 - 2\left( {1 - 2{{\sin }^2}\frac{t}{2}} \right)} = 3\sqrt {4{{\sin }^2}\frac{t}{2}} = 6\left| {\sin \frac{t}{2}} \right| = 6\sin \frac{t}{2}{\text{ }}\left( {0 \leqslant t \leqslant 2\pi } \right).$

$\displaystyle L = 6\int\limits_0^{2\pi } {\sin \frac{t}{2}dt} = 12\int\limits_0^{2\pi } {\sin \frac{t}{2}d\left( {\frac{t}{2}} \right)} = \left. { - 12\cos \frac{t}{2}} \right|_0^{2\pi } = - 12\left( { - 1 - 1} \right) = 24.$

Look this graph (cycloid)