Soution to Problem 301 | Definite Integral

Submitted by Anonymous (not verified) on Fri, 05/27/2011 - 13:06

Problem 301

যদি [tex]r = 2(1 - \cos \theta )[/tex] হয়, তবে দেখাও যে, [tex]\int\limits_0^\pi  {\sqrt {{r^2} + {{({{dr} \over {d\theta }})}^2}d\theta } }  = 8[/tex]

 

Answer:

যেহেতু, [tex]r = 2(1 - \cos \theta ) = 2.2{\sin ^2}{\theta  \over 2} = 4{\sin ^2}{\theta  \over 2}[/tex]

সুতরাং,  [tex]{{dr} \over {d\theta }} = 4.2\sin {\theta  \over 2}.{d \over {d\theta }}\sin {\theta  \over 2} = 4.2\sin {\theta  \over 2}.\cos {\theta  \over 2}.{1 \over 2} = 4\sin {\theta  \over 2}\cos {\theta  \over 2}[/tex]

 

অতএব [tex]{r^2} + {\left( {{{dr} \over {d\theta }}} \right)^2} = {\left( {4{{\sin }^2}{\theta  \over 2}} \right)^2} + {\left( {4\sin {\theta  \over 2}\cos {\theta  \over 2}} \right)^2}[/tex]

[tex] = 16{\sin ^2}{\theta  \over 2}\left( {{{\sin }^2}{\theta  \over 2} + {{\cos }^2}{\theta  \over 2}} \right) = 16{\sin ^2}{\theta  \over 2}[/tex]

 

অতএব [tex]\sqrt {{r^2} + {{\left( {{{dr} \over {d\theta }}} \right)}^2}}  = 4\sin {\theta  \over 2}[/tex]

 

অতএব [tex]\int\limits_0^\pi  {\sqrt {{r^2} + {{\left( {{{dr} \over {d\theta }}} \right)}^2}} d\theta  = } \int\limits_0^\pi  4 \sin {\theta  \over 2}d\theta  = 4\left[ { - {{\cos {\theta  \over 2}} \over {{1/2}}}} \right]_0^\pi[/tex]

[tex] =  - 8\left( {\cos {\pi  \over 2} - \cos 0} \right) =  - 8.( - 1) = 8[/tex] (প্রমাণিত)

 

 

 

 

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