Question

Find the area inside the cardioid r=5+4cos(θ)

Answer 1

Explanation:

Given:
`r=5+4cos{\theta}`

Area inside the cardioid is given by: A=
`\int\int\limits_D {r} \, drd{\theta}`

=
`\int_(0)^(2\pi)d\theta\int_(0)^(5+4cos\theta)rdr`

=
`(1)/(2)\int_(0)^(2\pi)d\theta(r^(2))_(0)^(5+4cos\theta)`

=
`(1)/(2)\int_(0)^(2\pi)(5+4cos\theta)^(2)d\theta`

=
`(1)/(2)\int_(0)^(2\pi)(25+16cos^2\theta+40cos\theta)`

=
`(1)/(2)\int_(0)^(2\pi)(25+8+8cos2\theta+40cos\theta)`

=
`(1)/(2)\int_(0)^(2\pi)(33+8cos2\theta+40cos\theta)`

=
`(1)/(2)(33\theta+16sin\theta+40sin\theta)_(0)^(2\pi)`

=
`(1)/(2)((33(2\pi))+16sin(2\pi))+40sin(2\pi))`

=
`33{\pi}`

Thus, the area inside the cardoid=
`33{\pi}`