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.)
Best Answer
The complication that arises when a polar curve is described by its "radius-squared" being given by a trigonometric function is that we may need to deal with negative and even imaginary radii. There is a simple protocol for working with "negative" radii, but having imaginary ones will introduce gaps in the angles covered by the curve.
In the case of the lemniscate described by $ \ r^2 \ = \ 162 \ \cos (2 \theta) \ $ , in the principal circle $ \ 0 \ \le \ \theta \ < \ 2 \pi \ $ , there are both positive and negative radii in the intervals $ \ 0 \ \le \ \theta \ \le \ \frac{\pi}{4} \ $ , $ \ \frac{3 \pi}{4} \ \le \ \theta \ \le \ \frac{5 \pi}{4} \ $ , and $ \ \frac{7 \pi}{4} \ \le \ \theta \ < \ 2 \pi \ $ . This has the effect of simultaneously "sweeping out" the half of one lobe that lies in the first quadrant and the half of the other lobe found in the third quadrant; these tracings of the curve meet at the origin for $ \ \theta \ = \ \frac{\pi}{4} \ $ . The "tracing" is then interrupted by having imaginary radii until $ \ \theta \ = \ \frac{3 \pi}{4} \ $ , where the other two half-lobes are then swept out up to $ \ \theta \ = \ \pi \ $ , completing the lemniscate. (Continuing further in angle simply re-traces the curve.)
The origin represents the angles $ \ \frac{(2k + 1) \ \pi}{4} \ , \ $ for all integers $ \ k \ $ .
To evaluate the area of the lemniscate, we can most easily avoid confusion by working with the portion of the right-hand lobe lying in quadrant I , which is traced over the interval $ \ 0 \ \le \ \theta \ \le \ \frac{\pi}{4} \ . $ We may integrate over this region and multiply the result by 4 :
$$ A \ = \ \ 4 \ \int_0^{\pi / 4} \ \frac{1}{2} [r(\theta)]^2 \ \ d\theta \ \ = \ \ 2 \ \int_0^{\pi / 4} \ 162 \ \cos (2 \theta) \ \ d\theta $$
$$ = \ 2 \cdot 162 \ \left[ \ \frac{1}{2} \ \sin (2 \theta) \ \right] \vert_0^{\pi / 4} \ = \ 162 \ ( \ \sin \ \frac{\pi}{2} \ - \ \sin \ 0 \ ) \ = \ 162 \ \ . $$
Indeed, we see that we can easily generalize this to say that the area of a lemniscate described by $ \ r^2 \ = \ C \ \cos (2 \theta) \ $ is just $ \ A \ = \ C \ $ .
[As a check on the credibility of our result, we can take the total area of the lemniscate as approximated by two ellipses with major axes of $ \ \sqrt{162} \ = \ 9\sqrt{2} \ $ . If we use the "height above" the $ \ x-$ axis at $ \ \theta \ = \ \frac{\pi}{6} \ $ as the semi-minor axis, this is $ \ [ \ \sqrt{162 \ \cos (2 \cdot \frac{\pi}{6})} \ ] \ \sin \frac{\pi}{6} \ = \ \frac{9}{2} \ $ . The area approximation by ellipses is then $ \ 2 \cdot \pi a b \ = \ 2 \ \cdot \ \pi \ \cdot \ \frac{9}{2} \sqrt{2} \ \cdot \ \frac{9}{2} \ = \ \frac{81 \sqrt{2}}{2} \pi \ \approx \ 180 \ $ , which agrees to about a 10% difference.]