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  1. #1
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    Application of integral calculus-3

    Please help me to solve this task
    Find a cardioid r = 1 - sin(fi) area except the part that lies on the circle r = sin(fi).

    Thanks in advance!
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  3. #3
    Senior Member DeMath's Avatar
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    {r_1} = 1 - \sin \varphi ,{\text{ }}{r_2} = \sin \varphi ,{\text{ }}{S_{{r_1}}}\backslash \left\{ {{S_{{r_1}}} \cap {S_{{r_2}}}} \right\} = ?

    If I understand the condition of your problem, my next decision

    S = {S_{{r_1}}}\backslash \left\{ {{S_{{r_1}}} \cap {S_{{r_2}}}} \right\} = {S_1} - 2\left( {{S_2} + {S_3}} \right) where

    {S_1} = \left\{ {\left. {\left( {\varphi ,r} \right)} \right|{\text{ 0}} \leqslant {r_1} \leqslant 1 - \sin \varphi ,{\text{ 0}} \leqslant \varphi  \leqslant 2\pi } \right\},

    {S_2} = \left\{ {\left. {\left( {\varphi ,r} \right)} \right|{\text{ 0}} \leqslant {r_1} \leqslant 1 - \sin \varphi ,{\text{ }}\frac{\pi }<br />
{6} \leqslant \varphi  \leqslant \frac{\pi }{2}} \right\},

    {S_3} = \left\{ {\left. {\left( {\varphi ,r} \right)} \right|{\text{ 0}} \leqslant {r_2} \leqslant \sin \varphi ,{\text{ 0}} \leqslant \varphi  \leqslant \frac{\pi }{6}} \right\}.

    S = \frac{1}{2}\int\limits_0^{2\pi } {r_1^2d\varphi }  - 2\left[ {\frac{1}{2}\int\limits_0^{\pi /6} {r_2^2d\varphi  + \frac{1}{2}\int\limits_{\pi /6}^{\pi /2} {r_1^2d\varphi } } } \right] =

    = \frac{1}{2}\int\limits_0^{2\pi } {{{\left( {1 - \sin \varphi } \right)}^2}d\varphi }  - \left[ {\int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}}{{{\sin }^2}\varphi d\varphi }  + \int\limits_{\pi /6}^{\pi /2} {{{\left( {1 - \sin \varphi } \right)}^2}d\varphi } } \right].

    \int {{{\left( {1 - \sin \varphi } \right)}^2}d\varphi }  = \int {\left( {1 - 2\sin \phi  + {{\sin }^2}\phi } \right)d\varphi }  =

    = \phi  + 2\cos \phi  + \frac{1}{2}\int {\left( {1 - \cos 2\phi } \right)d\phi }  = \frac{3}{2}\phi  + 2\cos \phi  - \frac{1}{4}\sin 2\phi  + C.

    {S_1} = \frac{1}{2}\int\limits_0^{2\pi } {{{\left( {1 - \sin \phi } \right)}^2}d\phi }  = \frac{1}{2} \left. {\left( {\frac{3}{2}\phi  + 2\cos \phi  - \frac{1}{4}\sin 2\phi } \right)} \right|_0^{2\pi } = \frac{3}{2}\pi .

    {S_2} = \int\limits_{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.<br />
 \kern-\nulldelimiterspace} 6}}^{{\pi  \mathord{\left/{\vphantom {\pi  2}} \right.\kern-\nulldelimiterspace} 2}} {{{\left( {1 - \sin \phi } \right)}^2}d\phi }  = \left. {\left( {\frac{3}{2}\phi  + 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  2}} \right.\kern-\nulldelimiterspace} 2}}<br />
= \frac{{3\pi }}{4} - \left( {\frac{{3\pi }}{{12}} + \sqrt 3  - \frac{{\sqrt 3 }}{8}} \right) = \frac{\pi }{2} - \frac{{7\sqrt 3 }}{8}.

    {S_3} = \int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} {{{\sin }^2}\varphi d\varphi }  = \frac{1}{2}\int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.<br />
 \kern-\nulldelimiterspace} 6}} {\left( {1 - \cos 2\phi } \right)d\varphi }  = \frac{1}{4} \left. {\left( {2\phi  - \sin 2\phi } \right)} \right|_0^{{\pi  \mathord{\left/{\vphantom {\pi  6}} \right.\kern-\nulldelimiterspace} 6}} = \frac{1}{4}\left( {\frac{\pi }{3} - \frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{{12}} - \frac{{\sqrt 3 }}{8}.

    S = \frac{3}{2}\pi  - \left( {\frac{\pi }{2} - \frac{{7\sqrt 3 }}{8} + \frac{\pi }{{12}} - \frac{{\sqrt 3 }}{8}} \right) = \frac{3}{2}\pi  - \frac{7}<br />
{{12}}\pi  + \sqrt 3  = \frac{{11}}{{12}}\pi  + \sqrt 3 .

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