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Math Help - Application of integral calculus - 2

  1. #1
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    Application of integral calculus - 2

    Please, help me

    1. Find a square which limited the polar curve r = 7(1- sin(fi)), -pi/6 =< fi =< pi/6.
    2. Find a length of the curve y = 2 + ln(x), sqrt(3) =< x =< sqrt(8).

    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!
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Dext91 View Post
    Please, help me

    1. Find a square which limited the polar curve r = 7(1- sin(fi)), -pi/6 =< fi =< pi/6.
    2. Find a length of the curve y = 2 + ln(x), sqrt(3) =< x =< sqrt(8).

    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!
    For number 1

    So r = 7\left( {1 - \sin \phi } \right),{\text{ }} - \frac{\pi }{6} \leqslant \phi  \leqslant \frac{\pi }{6},{\text{ }}S = ?

    Solution:

    S = \frac{1}<br />
{2}\int\limits_\alpha ^\beta  {{r^2}d\phi }  = \frac{1}<br />
{2}\int\limits_{{{ - \pi } \mathord{\left/<br />
 {\vphantom {{ - \pi } 6}} \right.<br />
 \kern-\nulldelimiterspace} 6}}^{{\pi  \mathord{\left/<br />
 {\vphantom {\pi  6}} \right.<br />
 \kern-\nulldelimiterspace} 6}} {{{\left( {7\left( {1 - \sin \phi } \right)} \right)}^2}d\phi }  =

     = \frac{{49}}<br />
{2}\int\limits_{{{ - \pi } \mathord{\left/<br />
 {\vphantom {{ - \pi } 6}} \right.<br />
 \kern-\nulldelimiterspace} 6}}^{{\pi  \mathord{\left/<br />
 {\vphantom {\pi  6}} \right.<br />
 \kern-\nulldelimiterspace} 6}} {\left( {1 - 2\sin \phi  + {{\sin }^2}\varphi } \right)d\phi }  =

    = \left. {\frac{{49}}{2}\left( {\frac{{3\phi }}{2} + 2\cos \phi } \right)} \right|_{{{ - \pi } \mathord{\left/{\vphantom {{ - \pi } 6}} \right.\kern-\nulldelimiterspace} 6}}^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} - \frac{{49}}<br />
{4}\int\limits_{{{ - \pi } \mathord{\left/<br />
 {\vphantom {{ - \pi } 6}} \right.<br />
 \kern-\nulldelimiterspace} 6}}^{{\pi  \mathord{\left/<br />
 {\vphantom {\pi  6}} \right.<br />
 \kern-\nulldelimiterspace} 6}} {\cos 2\phi d\phi }  =


    = \left. {\frac{{49}}{2}\left( {\frac{{3\phi }}{2} + 2\cos \phi  - \frac{1}{4}\sin 2\phi } \right)} \right|_{{{ - \pi } \mathord{\left/{\vphantom {{ - \pi } 6}} \right.\kern-\nulldelimiterspace} 6}}^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} = \frac{{49}}{2}\left( {2\pi  - \sqrt 3 } \right).

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  3. #3
    MHF Contributor Calculus26's Avatar
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    Answer to second question

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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Dext91 View Post
    Please, help me

    1. Find a square which limited the polar curve r = 7(1- sin(fi)), -pi/6 =< fi =< pi/6.
    2. Find a length of the curve y = 2 + ln(x), sqrt(3) =< x =< sqrt(8).

    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!
    So y = 2 + \ln x,{\text{ }}\sqrt 3  \leqslant x \leqslant \sqrt 8 ,{\text{ }}L = ?

    Solution:

    L = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}<br />
{{dx}}} \right)}^2}} dx}  = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}\left( {2 + \ln x} \right)} \right)}^2}} dx}  = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\sqrt {1 + \frac{1}{{{x^2}}}} dx}  =

    = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{1}{x}\sqrt {{x^2} + 1} dx}  = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{{{x^2} + 1}}{{x\sqrt {{x^2} + 1} }}dx}  = \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{x}<br />
{{\sqrt {{x^2} + 1} }}dx}  + \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{1}<br />
{{x\sqrt {{x^2} + 1} }}dx}  =

    = \int\limits_{\sqrt 3 }^{\sqrt 8 } {d\left( {\sqrt {{x^2} + 1} } \right) + } \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{x}{{{x^2}\sqrt {{x^2} + 1} }}dx}  = \left. {\sqrt {{x^2} + 1} } \right|_{\sqrt 3 }^{\sqrt 8 } + \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{x}{{\left( {{x^2} + 1 - 1} \right)\sqrt {{x^2} + 1} }}dx}  =

    = 1 + \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{{{x \mathord{\left/{\vphantom {x {\sqrt {{{\left( {{x^2} + 1} \right)}^3}} }}} \right.\kern-\nulldelimiterspace} {\sqrt {{{\left( {{x^2} + 1} \right)}^3}} }}}}<br />
{{1 - {{\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}^2}}}dx}  = 1 - \int\limits_{\sqrt 3 }^{\sqrt 8 } {\frac{{d\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}}{{1 - {{\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}^2}}}} =1 - \left. {\frac{1}{2}\ln \left| {\frac{{1 + \frac{1}{{\sqrt {{x^2} + 1} }}}}{{1 - \frac{1}{{\sqrt {{x^2} + 1} }}}}} \right|} \right|_{\sqrt 3 }^{\sqrt 8 }=

    = 1 - \left. {\frac{1}{2}\ln \left| {\frac{{\sqrt {{x^2} + 1}  + 1}}{{\sqrt {{x^2} + 1}  - 1}}} \right|} \right|_{\sqrt 3 }^{\sqrt 8 } = 1 - \frac{1}{2}\left( {\ln 2 - \ln 3} \right) = 1 + \frac{1}{2}\ln \frac{3}{2}.

    Look this graph

    Last edited by DeMath; July 17th 2009 at 08:40 PM.
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