# Application of integral calculus-3

• Jul 18th 2009, 06:48 AM
Dext91
Application of integral calculus-3
Find a cardioid r = 1 - sin(fi) area except the part that lies on the circle r = sin(fi).

• Jul 18th 2009, 07:38 AM
Calculus26
See attachment
• Jul 18th 2009, 08:13 AM
DeMath
$\displaystyle {r_1} = 1 - \sin \varphi ,{\text{ }}{r_2} = \sin \varphi ,{\text{ }}{S_{{r_1}}}\backslash \left\{ {{S_{{r_1}}} \cap {S_{{r_2}}}} \right\} = ?$

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

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

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

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

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

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

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

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

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

$\displaystyle {S_1} = \frac{1}{2}\int\limits_0^{2\pi } {{{\left( {1 - \sin \phi } \right)}^2}d\phi } = \frac{1}{2}$$\displaystyle \left. {\left( {\frac{3}{2}\phi + 2\cos \phi - \frac{1}{4}\sin 2\phi } \right)} \right|_0^{2\pi } = \frac{3}{2}\pi . \displaystyle {S_2} = \int\limits_{{\pi \mathord{\left/{\vphantom {\pi 6}} \right. \kern-\nulldelimiterspace} 6}}^{{\pi \mathord{\left/{\vphantom {\pi 2}} \right.\kern-\nulldelimiterspace} 2}} {{{\left( {1 - \sin \phi } \right)}^2}d\phi } =$$\displaystyle \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}} $$\displaystyle = \frac{{3\pi }}{4} - \left( {\frac{{3\pi }}{{12}} + \sqrt 3 - \frac{{\sqrt 3 }}{8}} \right) = \frac{\pi }{2} - \frac{{7\sqrt 3 }}{8}. \displaystyle {S_3} = \int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right.\kern-\nulldelimiterspace} 6}} {{{\sin }^2}\varphi d\varphi } = \frac{1}{2}\int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 6}} \right. \kern-\nulldelimiterspace} 6}} {\left( {1 - \cos 2\phi } \right)d\varphi } = \frac{1}{4}$$\displaystyle \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}.$

$\displaystyle S = \frac{3}{2}\pi - \left( {\frac{\pi }{2} - \frac{{7\sqrt 3 }}{8} + \frac{\pi }{{12}} - \frac{{\sqrt 3 }}{8}} \right) = \frac{3}{2}\pi - \frac{7} {{12}}\pi + \sqrt 3 = \frac{{11}}{{12}}\pi + \sqrt 3 .$

Look this picture