Finding the area of the region with double integrals

I have to find the area of the region inside $r^2=16\cos(2\theta)$ and inside $r=2\cos(\theta)$. Should I divide the positive x and y region into two parts? Or can I bound r by $\sqrt{16\cos(2\theta)}\ \text{and}\ 2\cos(\theta)$. I know that $\{\theta \mid 0<\theta<\frac{\pi}{4}\}$ for the positive x and y region. And once I find the area for y>0 and x>0,its only by symmetry that we multiply by 2 to find the complete region.

• Here is a similar question. – Shuchang Dec 11 '13 at 12:27
• I don't think its the same question. And the answer of that question n doesn't answer my question :( – John Dec 11 '13 at 12:31
• It really is preferable to make the area computation using polar coordinates, since the lemniscate has a less than simple Cartesian equation. – colormegone Apr 3 '14 at 6:52

This polar area integration does have a couple of bits that require careful handling, in part because of the character of the two curves, and partly because one of the limits of integration is not a "fundamental angle".

The curve $\ r \ = \ 2 \cos \theta \$ is a "one-petal" rosette, the sort that looks like a circle tangent to the origin with its center on the $\ x-$ axis. What is special about rosettes with an odd number of petals is that the curves are completely "swept out" with a period of $\ \pi \$ , rather than $\ 2 \pi \ .$ The other curve, $\ r^2 \ = \ 16 \ \cos (2\theta) \ ,$ is a lemniscate, which has the peculiar property that its polar equation produces non-real radii for half of its period .

We wish to find the area of the region within both curves (shown in green). Since it is symmetrical about the $\ x-$ axis, we will just integrate over the half above that axis and double the result. We need to know how the angle-variable "runs" along each curve, and the value of $\ \theta \$ at which the two curves intersect.

Both curves intercept the $\ x-$ axis at $\ \theta = 0 \ ,$ but each first reaches the origin at distinct values, the rosette at $\ \theta = \frac{\pi}{2} \ ,$ the lemniscate at $\ \theta = \frac{\pi}{4} \ .$ Since we will be working in the first quadrant, it is "safe" to solve for the intersection point of the curves by equating $\ r^2 \$ for the two: this will not introduce "spurious" solutions. We obtain

$$16 \ \cos 2 \theta \ = \ 4 \ \cos^2 \theta \ \ \Rightarrow \ \ 4 \ \cos 2 \theta \ = \ \frac{1}{2} ( \ 1 \ + \ \cos 2\theta \ )$$

$$\Rightarrow \ \ 8 \ \cos 2 \theta \ = \ 1 \ + \ \cos 2\theta \ \ \ \Rightarrow \ \ \cos 2 \theta \ = \ \frac{1}{7} \ \ ,$$

employing familiar trigonometric identities. The solution for $\ \theta \$ is not any "fundamental angle" (it proves to be $\ \Theta \ \approx \ 0.7137 \$ , as a graph of the functions will verify), so we will actually prefer to work with our result for $\ \cos 2 \Theta \ .$ (We will shortly have need of the value $\ \sin 2 \Theta \ = \ \frac{\sqrt{48}}{7} \ .$ )

The "upper half" of the area of the region is covered then by integrating the area within the rosette from $\ \theta = 0 \ \ \text{to} \ \ \theta = \Theta \ ,$ and then passing over to the lemniscate from $\ \theta = \Theta \ \ \text{to} \ \ \theta = \frac{\pi}{4} \ .$ (This does behave properly, since $\ \Theta \ < \ \frac{\pi}{4} \ \approx \ 0.7854 \ .$ ) The area of the entire region is then found from

$$2 \ \left[ \ \int_0^{\Theta} \ \frac{1}{2} r^2_{ros} \ \ d\theta \ \ + \ \ \int^{\pi / 4}_{\Theta} \ \frac{1}{2} r^2_{lemn} \ \ d\theta \ \right]$$

$$= \ \ \int_0^{\Theta} \ ( \ 2 \cos \theta \ )^2 \ \ d\theta \ \ + \ \ \int^{\pi / 4}_{\Theta} ( \ 16 \ \cos 2\theta \ ) \ \ d\theta$$

$$= \ \ 4 \ \int_0^{\Theta} \cos^2 \theta \ \ d\theta \ \ + \ \ 16 \ \int^{\pi / 4}_{\Theta} \cos 2\theta \ \ d\theta$$

$$= \ \ 2 \ \int_0^{\Theta} ( \ 1 + \cos 2 \theta \ ) \ \ d\theta \ \ + \ \ 16 \ \int^{\pi / 4}_{\Theta} \cos 2\theta \ \ d\theta$$

$$= \ \ \left( \ 2 \theta \ + \ \sin 2 \theta \ \right) \vert_0^{\Theta} \ + \ \left( \ 8 \ \sin 2 \theta \ \right) \vert^{\pi / 4}_{\Theta}$$

$$= \ \ ( \ 2 \Theta \ + \ \sin 2 \Theta \ - \ 0 \ - \ 0) \ + \ ( \ 8 \ - \ 8 \ \sin 2 \Theta \ )$$

$$= \ \ 8 \ + \ 2 \Theta \ - \ 7 \ \sin 2 \Theta \ = \ 8 \ + \ \arccos \frac{1}{7} \ - \ 7 \ \cdot \frac{\sqrt{48}}{7}$$

$$= \ 8 \ + \ \arccos \frac{1}{7} \ - \ \sqrt{48} \ \approx \ 2.4992 \ \ .$$

The region lies within the rosette, which has area $\ \pi \ ,$ so this result is reasonable. (In fact, it fills very close to 80% of the rosette.)