Application of integral calculus. Find a length

**Dext91** · July 17th 2009, 07:20 AM

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.

Thank advance!

**DeMath** · July 17th 2009, 08:19 AM

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.

Thank advance!

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

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

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

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

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

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

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

Look this picture

**DeMath** · July 17th 2009, 08:23 AM

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.

Thank advance!

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

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

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

Look this picture

**DeMath** · July 17th 2009, 08:31 AM

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.
Thank advance!

So we have $r = a\cos 3\varphi ,{\text{ }}0 \leqslant \varphi \leqslant 2\pi ,{\text{ }}S - ?$

Solution

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

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

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

Look this picture

**DeMath** · July 17th 2009, 08:49 AM

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.
Thank advance!

$x = 3\left( {t - \sin t} \right),{\text{ }}y = 3\left( {1 - \cos t} \right),{\text{ }}0 \leqslant t \leqslant 2\pi ,{\text{ }}L = ?$

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

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

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

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

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

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