Question

Find the area bounded by the polar curves? `r=4+4cos theta` and `r=6`

Answer 1

Bounded Area = `18sqrt(3) - 4pi`

Explanation:

`r=4+4costheta \ \ \ \ \ \ (color(red)(red))`
`r=6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (color(blue)(blue))`

The area we seek is shaded in grey.

Let us first find the points of intersection:

`4+4costheta=6`
`:. 4cos theta = 2`
`:. cos theta = 1/2`
`:. theta = +-pi/3`

We calculate area in polar coordinates using :

`A = 1/2 \ int_alpha^beta \ r^2 \ d theta`

The area bounded by `r=4+4costheta` and `r=+-pi/3` is:

`A_1 = 1/2 \ int_(-pi/3)^(pi/3) \ (4+4costheta)^2 \ d theta`
`\ \ \ \ = 1/2 \ int_(-pi/3)^(pi/3) \ 16(1+costheta)^2 \ d theta`
`\ \ \ \ = 8 \ int_(-pi/3)^(pi/3) \ 1+2costheta+cos^2theta \ d theta`
`\ \ \ \ = 8 \ int_(-pi/3)^(pi/3) \ 1+2costheta+(1+cos2theta)/2 \ d theta`
`\ \ \ \ = 8 \ int_(-pi/3)^(pi/3) \ 1+2costheta+1/2+(cos2theta)/2 \ d theta`
`\ \ \ \ = 8 \ int_(-pi/3)^(pi/3) \ 3/2+2costheta+(cos2theta)/2 \ d theta`
`\ \ \ \ = 8 [3/2theta+2sintheta+(sin2theta)/4]_(-pi/3)^(pi/3)`

`\ \ \ \ = 8 { (3/2(pi/3) + 2sin(pi/3)+1/4sin((2pi)/3)) - (3/2(-pi/3) + 2sin(-pi/3)+1/4sin((-2pi)/3))}`

`\ \ \ \ = 8 { ( pi/2 + 2sqrt(3)/2+1/4sqrt(3)/2) - (-pi/2 - 2sqrt(3)/2-1/4sqrt(3)/2)}`

`\ \ \ \ = 8 { pi/2 + 2sqrt(3)/2+1/4sqrt(3)/2 +pi/2 + 2sqrt(3)/2+1/4sqrt(3)/2}`

`\ \ \ \ = 8 { pi + 2sqrt(3)+1/4sqrt(3)}`

`\ \ \ \ = 8pi + 18sqrt(3)`

The area bounded by `r=6` and `r=+-pi/3` is a sector of angle `(2pi)/3` so we can use the sector area `1/2r^2theta`

`A_2 = 1/2(6^2)(2pi)/3`
`\ \ \ \ = 1/2 xx 36 xx (2pi)/3`
`\ \ \ \ = 12pi`

Then, the difference is:

`A = A_1 - A_2`
`\ \ \ = (8pi + 18sqrt(3)) - (12pi)`
`\ \ \ = 18sqrt(3) - 4pi`

Note:

We could also construct a single integral to get the solution using:

`A = 1/2 \ int_(-pi/3)^(pi/3) \ (4+4costheta)^2 - (6)^2 \ d theta`