Question

Find the area bounded by the inside of the polar curve `r=1+cos 2theta` and outside the polar curve `r=1`?

Answer 1

`2+pi/4`

Explanation:

Here is the graph of the two curves. The shaded area, `A`, is the area of interest:

It is a symmetrical problems so we only need find the shaded area of the RHS of Quadrant `1` and multiply by `4`.

We could find the angle `theta` in `Q1` for the point of interaction by solving the simultaneous equations:

`r=1+cos 2theta`
`r=1`

However, intuition is faster, and it looks like angle of intersection in `Q1` is `pi/4`, we can verify this by a quick evaluation:

`theta = pi/4 => r=1+cos 2theta =1+cos(pi/2) = 1`

Confirming our intuition. So now we have the following:

Where the shaded area repents `1/4` of the total area sought:

We can now start to set up a double integral to calculate this area

`r` sweeps out a ray from `1` to `1+cos(2theta)`
`theta` varies for `0` to `pi/4`

So then:

`1/4A = int_0^(pi/4) int_1^(1+cos2theta) r \ dr \ d theta`

If we evaluate the inner integral first then we get:

`int_1^(1+cos2theta) r \ dr = [1/2r^2]_1^(1+cos2theta)`
`" " = 1/2( (1+cos2theta)^2 - 1^2)`
`" " = 1/2( 1+2cos2theta+cos^2 2theta - 1)`
`" " = 1/2( 2cos2theta+cos^2 2theta)`
`" " = 1/2( 2cos2theta+1/2(cos4theta+1))`
`" " = cos2theta+1/4cos4theta+1/4`

And so our double integral becomes:

`1/4A \ = int_0^(pi/4) int_1^(1+cos2theta) r \ dr \ d theta`
`" " = int_0^(pi/4) cos2theta+1/4cos4theta+1/4 \ d theta`
`" " = [1/2sin2theta+1/16sin4theta+1/4theta]_0^(pi/4)`
`" " = (1/2sin(pi/2)+1/16sinpi+1/4pi/4) - (0)`
`" " = 1/2+pi/16`
`:. A = 2+pi/4`

``

METHOD 2

If you are not happy with double integrals, then we can evaluate the area using the polar area formula `A=int \ 1/2r^2 \ d theta` and calculating the two shaded sections separately:

Here this shaded area is given by:

`A_1 = 1/2 \ int_0^(pi/4) \ (1+cos 2theta)^2 \ d theta`
`\ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+ cos^2 2theta \ d theta`
`\ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+1/2(cos4theta+1) \ d theta`
`\ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+1/2cos4theta+1/2 \ d theta`
`\ \ \ \ = 1/2 [3/2theta+sin2theta+1/8sin4theta]_0^(pi/4)`
`\ \ \ \ = 1/2 ((3pi)/8+1+0)`
`\ \ \ \ = (3pi)/16+1/2`

And this shaded area

is given by:

`A_2 = 1/2 \ int_0^(pi/4) \ (1)^2 \ d theta`
`\ \ \ \ = 1/2 [theta]_0^(pi/4)`
`\ \ \ \ = 1/2 (pi/4-0)`
`\ \ \ \ = pi/8`

And so the total sought area is `4` times the difference between these results:

`1/4 A \ = A_1 - A_2`
`" " = (3pi)/16+1/2 - pi/8`
`" " = (pi)/16+1/2`
`:. A = (pi)/4+2`

Answer 2

`= 2 + pi/4`

Explanation:

First a serious time saver which is a plot.

We are asked for the area hatched in black. The intersection points labelled a - d are solutions to:

`1 = 1 + cos 2 theta implies cos 2 theta = 0`

`implies theta = - pi/4, pi/4, (3 pi)/4, (5 pi)/4 ,....`

The four angles listed above are those marked a - d on the plot.

Next, by my reckoning an important thing to understand - how polar areas are calculated. There is a simple formula for the area of polar function `r(theta)` in the interval `theta in [alpha, beta]`:

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

There are various ways to get to this but the geometric (hand-wavey) derivation shows what it actually means:

In this drawing, the area is the green section, which is "swept out" between the angles `alpha` and `beta` as shown.

This means that, in our case, the area in the next illustration is:

`A = 1/2 int_(-pi/4)^(pi/4) (1 + cos 2 theta )^2 d theta`

Clearly that includes some of the circle that is outside what we are trying to evaluate. In addition, it is only half the story. If you explore `r(-theta)` and `r(pi - theta)` you will see that this is symmetric about both Cartesian axes. We can therefore focus only on the area in Q1:

`A = 1/2 int_(color(red)(0))^(pi/4) (1 + cos 2 theta )^2 d theta`

From that we must subtract the area of `1/8`th of the unit circle. Recap that intersection point "b" is t `theta = pi/4`.

So our modified Area is:

`A = 1/2 int_(color(red)(0))^(pi/4) (1 + cos 2 theta )^2 d theta color(red)(- pi/8)`

But we have four of these so our final answer is:

`A =4 ( 1/2 int_(color(red)(0))^(pi/4) (1 + cos 2 theta )^2 d theta - pi/8 ) = 2 + pi/4`

(I've assumed you know how to manage that integrand.)