Question

Using Polar coordinates, what is the area of region A?

F: `r` = 16 `cos(Theta)`
`r_f` = 8

G: `r` = 5
`r_g` = 5

Answer 1

`( (5/2)sqrt 231 + 103 cos^(-1)(5/16) )`
`= ( 5/2 sqrt 231 + 129.056 rad)`= 167.05 areal units, nearly.
This is nearly 83.1 % of the area of the larger circle.

Explanation:

At the common points,

`r = 16 cos theta = 5`, giving `cos theta = 5/16` and

`(5, +-cos^(-1)(5/16))`, as the points of intersection.

So, the shaded area

`= 2 int [int r dr] d theta`,

between the limits 5 and `16 costheta`, for r,

and 0 and `cos^(-1)(5/16)`, for `theta`

`= int (16^2cos^2theta - 5^2) d theta`,

`theta` from 0 to `cos^(-1) (5/16)`

`= int (256 cos^2theta -25) d theta, theta` from 0 to #cos^(-1)

(5/16)#

`= int ( 128 cos 2theta + 103 )] d theta`, between the limits

`=[ 64 sin 2theta + 103 theta ]`, between the limits

`=( 128( sqrt( 1-(5/16)^2)(5/16)) + 103 cos(-1)(5/16))`

`= ( (5/2)sqrt 231 + 103cos(-1)(5/16) )`

`= ( 5/2 sqrt 231 + 129.056 rad)`

#= 167.05 areal units, nearly.