Question

The area common to the cardioids $r=a(1+\cos \theta )$ and $r=a(1-\cos \theta )$ is:

• A. $(\frac{3\pi }{2}+4)a^2$
• B. $(\frac{3\pi }{2}-4)a^2$
• C. $(\frac{\pi }{2}+4)a^2$
• D. $(\frac{\pi }{2}-4)a^2$

Answer 1

The correct option is B $(\frac{3\pi }{2}-4)a^2$

The cardioid $r=a(1+\cos \theta )$ is $ABCO{B}^{\prime }A$ and the cardioid $r=a(1-\cos \theta )$ is $O{C}^{\prime }B{A}^{\prime }{B}^{\prime }O$
Both the cardioids are symmetrical about the initial line $OX$ and intersect at $B$ and $B^{\prime }$
$\therefore$ Required Area$=2$Area $O{C}^{\prime }BCO$
$=2$[area$O{C}^{\prime }BO+$area$OBCO$]
$=2[{\left({\int }_0^{\frac{\pi }{2}}\frac{1}{2}{r}^{2}d\theta \right)}_{r=a(1-\cos \theta )}+{\int }_{\frac{\pi }{2}}^{\pi }{\left({(1+\cos \theta )}^{2}d\theta \right)}_{r=a(1+\cos \theta )}]$
$=a^2[{\int }_0^{\frac{\pi }{2}}(1-2\cos \theta +{\cos }^2\theta )d\theta +{\int }_{\frac{\pi }{2}}^{\pi }(1+2\cos \theta +{\cos }^2\theta )d\theta ]$
$=a^2[{\int }_0^{\pi }(1+{\cos }^2\theta )d\theta -2{\int }_0^{\frac{\pi }{2}}\cos \theta d\theta +2{\int }_{\frac{\pi }{2}}^{\pi }\cos \theta d\theta ]$
$=a^2[{\int }_0^{\pi }(1+\frac{1+\cos 2\theta }{2})d\theta -2{\left|\sin \theta \right|}_0^{\frac{\pi }{2}}+2{\left|\sin \theta \right|}_{\frac{\pi }{2}}^{\pi }]$
$=a^2[{|\frac{3}{2}\theta +\frac{\sin 2\theta }{4}|}_0^{\pi }-2(1-0)+2(0-1)]$
$=(\frac{3\pi }{2}-4)a^2$