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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.

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

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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.

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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.)

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