# How do we find the area inside the cardioid f(theta)=a(1+costheta)?

May 27, 2016

Area of the region inside cardioid is $\frac{3 \pi {a}^{2}}{2}$

#### Explanation:

Area inside the cardioid or any other bound curve is found using integral

${\int}_{0}^{2 \pi} \frac{1}{2} {\left(f \left(\theta\right)\right)}^{2} d \theta$

As here $f \left(\theta\right) = a \left(1 + \cos \theta\right)$

Area is ${\int}_{0}^{2 \pi} \frac{1}{2} {\left(a \left(1 + \cos \theta\right)\right)}^{2} d \theta$

= ${a}^{2} / 2 {\int}_{0}^{2 \pi} {\left(1 + \cos \theta\right)}^{2} d \theta$

= ${a}^{2} / 2 {\int}_{0}^{2 \pi} \left(1 + 2 \cos \theta + {\cos}^{2} \theta\right) d \theta$

= ${a}^{2} / 2 | \theta + 2 \sin \theta {|}_{0}^{2 \pi} + {a}^{2} / 2 {\int}_{0}^{2 \pi} {\cos}^{2} \theta d \theta$ (A)

Now ${\cos}^{2} \theta = \frac{1}{2} \left(1 + \cos 2 \theta\right)$

Hence ${\int}_{0}^{2 \pi} {\cos}^{2} \theta d \theta = {\int}_{0}^{2 \pi} \left(\frac{1}{2} + \frac{1}{2} \cos 2 \theta\right) d \theta$

= $| \frac{\theta}{2} + \frac{1}{4} \sin 2 \theta {|}_{0}^{2 \pi}$ (B)

Hence, substituting (B) in (A)

${\int}_{0}^{2 \pi} \frac{1}{2} {\left(a \left(1 + \cos \theta\right)\right)}^{2} d \theta$

= ${a}^{2} / 2 | \theta + 2 \sin \theta + \frac{\theta}{2} + \frac{1}{4} \sin 2 \theta {|}_{0}^{2 \pi}$

= ${a}^{2} / 2 \left(2 \pi + 2 \sin \left(2 \pi\right) + \frac{2 \pi}{2} + \frac{1}{4} \sin \left(4 \pi\right) - 0 - 2 \sin 0 - \frac{0}{2} - \frac{1}{4} \sin \left(4 \pi\right)\right)$

= ${a}^{2} / 2 \times 3 \pi = \frac{3 \pi {a}^{2}}{2}$