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.
[Math] Finding the area of the region with double integrals
areacalculusmultivariable-calculus
Related Solutions
There isn't much difference between doing area integration in polar coordinates as a double integral and in the way you may have encountered it earlier in single-variable calculus. It is still important to have an idea of what the regions look like (here, you have a limacon and a "peanut").
Your radial portion is correct, but you really need to look at the angles where the curves intersect: you'll want to solve $ \ 2 + \sin \theta \ = \ 2 + \cos (2 \theta)$ to get the range of angle integration. There are two zones to cover, but you can make use of symmetry here and just integrate over one of them.
The red curve is the limacon $2 + \sin\theta$ , the blue curve, $2 + \cos(2 \theta)$ .
Incidentally, in the integral you wrote, it's not that there is no integrand, but rather that the integrand is "1". You are doing a surface integral where all infinitesimal patches are equally "weighted" at 1 , so the result is simply the area of the region. Surface integrals with some "weighting function" $f(r,\theta)$ or $g(x,y)$ turn up in various applications.
Best Answer
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.)