Area of the three-petal rose $r=2\cos 3\theta$. Why does $\frac12 \int_0^{2\pi} r^2d\theta$ give the wrong result? For the polar curve $$r=2\cos(3\theta)$$, you can find the area of all three petals by getting the area of one-half petal using the bounds $$\theta = 0$$ to $$\theta = \pi$$/6 for $$A = (1/2)\int r^2 d\theta$$, and multiplying that by 6. But when you try to get the area of the curve using the bounds $$\theta = 0$$ to $$\theta = 2\pi$$, the area is incorrect. My reasoning for why the latter method should work is that $$A = (1/2)\int r^2 d \theta$$, which will be the area within the bounds set by the integral. So why would the integral with the bounds $$0$$ to $$2\pi$$ not yield the correct area? The graph in the picture is for $$r=\cos(3\theta)$$.

While you haven't shown your computation, if you did it correctly you should have found that your integral from $$0$$ to $$2\pi$$ came out as twice the correct answer. The reason is that as $$\theta$$ increases from $$0$$ to $$2\pi$$, the rose curve is drawn twice. So you should only integrate from $$0$$ to $$\pi$$ to get the correct value. Note that $$\pi$$ is $$6\cdot \pi/6$$-this isn't coincidental. Each $$\pi/6$$ increase in $$\theta$$ draws half of a petal.
The reason is because when $$\cos 3\theta < 0$$, then the radius becomes negative. On the interval $$\theta \in [0,2\pi)$$, this occurs when $$\theta \in (\pi/6, \pi/2) \cup (5\pi/6, 7\pi/6) \cup (3\pi/2, 11\pi/6).$$ So in fact, this curve is traversed twice as $$\theta \in [0, 2\pi)$$.
Refer to the following diagram. As $$\theta$$ traverses the interval, when $$r > 0$$, the radial line with angle $$\theta$$ is blue and the point it is drawing is also blue. When $$r < 0$$, the line and point becomes red. The black circle represents the angle of the line, so as you can see, the line only makes a single complete rotation, but the curve is drawn twice. 