Question

What is the area bounded by the the inside of polar curve `1+cos theta` and outside the polar curve `r(1+cos theta)=1`?

Answer 1

`"Area" = (3pi)/4+ 4/3`

Explanation:

If we plot the polar curve, and shadethe area sought:

We observe that the points of intersection are:

`theta = -pi/2` and `theta=pi/2`

We calculate area in polar coordinates using :

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

Thus, the enclosed area is:

`A = 1/2 \ int_(-pi/2)^(pi/2) \ (1+cos theta)^2 - (1/(1+cos theta))^2 \ d theta`

`:. 2A = int_(-pi/2)^(pi/2) \ (1+cos theta)^2 \ d theta - int_(-pi/2)^(pi/2) (1/(1+cos theta))^2 \ d theta`

Consider the first integral;:

`I_1= int_(-pi/2)^(pi/2) \ (1+cos theta)^2 \ d theta`
`\ \ \ = int_(-pi/2)^(pi/2) \ 1+ 2cos theta + cos^2 theta \ d theta`
`\ \ \ = int_(-pi/2)^(pi/2) \ 1+ 2cos theta + (1+cos 2theta)/2 \ d theta`
`\ \ \ = int_(-pi/2)^(pi/2) \ 3/2+ 2cos theta + (cos 2theta)/2 \ d theta`
`\ \ \ = [(3theta)/2+ 2sin theta + (sin 2theta)/4]_(-pi/2)^(pi/2) \ `

`\ \ \ = ((3(pi/2))/2+ 2sin (pi/2)+ (sin (2pi))/4) - ((3(-pi/2))/2+ 2sin (-pi/2)+ (sin (-2pi))/4)`

`\ \ \ = ((3pi)/4+ 2+ 0) - (-(3pi)/4-2-0)`
`\ \ \ = (3pi)/2+ 4`

And, now the second integral:

`I_2 = int_(-pi/2)^(pi/2) (1/(1+cos theta))^2 \ d theta`

For which, we perform, a tangent half-angle substitution:

`u =tan(theta/2) => (du)/(d theta) = 1/2sec^2(theta/2)`

When:

`theta = { (pi/2), (-pi/2) :} => u = { (1), (-1) :}`

And we can manipulate the integral and perform the substitution, to get:

`I_2 = int_(-pi/2)^(pi/2) 1/( (1-tan^2(theta/2))/(1+tan^2(theta/2))+1)^2 \ d theta`
`\ \ \ = int_(-pi/2)^(pi/2) 1/( (1-tan^2(theta/2))/(sec^2(theta/2))+1)^2 \ d theta`
`\ \ \ = int_(-pi/2)^(pi/2) 1/ ((1-tan^2(theta/2)+1+tan^2(theta/2))/(1+tan^2(theta/2)))^2 \ d theta`
`\ \ \ = int_(-pi/2)^(pi/2) 1/ ((2)/(sec^2(theta/2)))^2 \ d theta`
`\ \ \ = int_(-pi/2)^(pi/2) 1/ ((2)/(sec^2(theta/2)))^2 \ d theta`
`\ \ \ = 1/4 \ int_(-pi/2)^(pi/2) sec^2 (theta/2) sec^2 (theta/2) \ d theta`
`\ \ \ = 1/2 \ int_(-pi/2)^(pi/2) (1+tan^2 (theta/2)) \ 1/2sec^2 (theta/2) \ d theta`
`\ \ \ = 1/2 \ int_(-1)^(1) 1+u^2 \ du`
`\ \ \ = 1/2 [u+u^3/3]_(-1)^(1)`
`\ \ \ = 1/2 {(1+1/3) - (-1-1/3)}`
`\ \ \ = 4/3`

Combining both results, we get:

`2A = I_1 + I_2`
`\ \ \ \ \ = (3pi)/2+ 4 + 4/3`
`\ \ \ \ \ = (3pi)/2+ 16/3`

Hence:

`A = (3pi)/4+ 4/3`