Answer:
Step-by-step explanation:
Given: [tex]r=5+4cos{\theta}[/tex]
Area inside the cardioid is given by: A=[tex]\int\int\limits_D {r} \, drd{\theta}[/tex]
=[tex]\int_{0}^{2\pi}d\theta\int_{0}^{5+4cos\theta}rdr[/tex]
=[tex]\frac{1}{2}\int_{0}^{2\pi}d\theta(r^{2})_{0}^{5+4cos\theta}[/tex]
=[tex]\frac{1}{2}\int_{0}^{2\pi}(5+4cos\theta)^{2}d\theta[/tex]
=[tex]\frac{1}{2}\int_{0}^{2\pi}(25+16cos^2\theta+40cos\theta)[/tex]
=[tex]\frac{1}{2}\int_{0}^{2\pi}(25+8+8cos2\theta+40cos\theta)[/tex]
=[tex]\frac{1}{2}\int_{0}^{2\pi}(33+8cos2\theta+40cos\theta)[/tex]
=[tex]\frac{1}{2}(33\theta+16sin\theta+40sin\theta)_{0}^{2\pi}[/tex]
=[tex]\frac{1}{2}((33(2\pi))+16sin(2\pi))+40sin(2\pi))[/tex]
=[tex]33{\pi}[/tex]
Thus, the area inside the cardoid= [tex]33{\pi}[/tex]
