Question

Find the area of the region inside these curves?

curves:

• `r=1+\cos\theta` (red)
• `r=1-\cos\theta` (blue)

Answer 1

`(3pi)/2 - 4`

Explanation:

The question is ambiguous, asking us to find the area inside the polar curves:

`r = 1 + cos theta` ..... [A]
`r = 1 - cos theta` ..... [B]

Assuming that we seek the area bounded by both curves:

First we test for symmetry (which would appear obvious) about `Ox` by replacing `theta` with `-theta` in [A] so we get:

`r = 1 + cos theta |-> r=1 + cos -theta = 1 - cos theta`

which is the equation of [B]. Similarly we test for symmetry about `Oy` by replacing `theta` with `pi-theta` in [A] so we get:

`r = 1 + cos theta |-> r=1 + cos (pi-theta)`
`=> r = 1 + cos pi cos theta + sin pi sin theta = 1 - cos theta`
which, again, is the equation of [B].

These test confirm that we have symmetry about `Ox` and `Oy`, thus we need only consider the area `A_1`, of a half petal of `r = 1 - cos theta` for `theta in [0, pi/2]`, and multiply by `4`

We calculate area in polar coordinates using :

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

Thus, the enclosed area of a single half petal is:

`A_1 = 1/2 \ int_(0)^(pi/2) \ (1 - cos theta)^2 \ d theta`

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

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

`\ \ \ \ = 1/2 \ int_(0)^(pi/2) \ 3/2 - 2cos theta + 1/2cos 2theta \ d theta`

`\ \ \ \ = 1/2 \ [ \ (3 theta)/2 - 2sin theta + 1/4sin 2theta \ ]_(0)^(pi/2)`

`\ \ \ \ = 1/2 \ {(3/2*pi/2 - 2sin(pi/2)+1/4sin(pi))-(0-2sin0+1/4sin0) }`

`\ \ \ \ = 1/2 \ {(3pi)/4 - 2+0)0(0)}`

So that the total area of all 4 half-petals, is:

`A = 4A_1`

`\ \ = 4 * 1/2 \ ((3pi)/4 - 2)`

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