This is the question I was given. I feel like it is supposed to be basic, but I am getting stuck. To me, it looks like if I find the area of half the red petal and subtract out the area of the part of the blue petal that is in the inside, I will get the correct answer. That gives me: $$\text{Area of half of red petal:}\frac12\int^{\frac{\pi}{6}}_04\cos^2(3\theta)\ d\theta \tag{1}$$$$\text{Area of section of blue petal:}\frac12\int^{\frac{\pi}{12}}_04\sin^2(3\theta)\ d\theta \tag2$$$$\text{Subtract second from first:}\frac12\int^{\frac{\pi}{6}}_04\cos^2(3\theta)\ d\theta -\frac12\int^{\frac{\pi}{12}}_04\sin^2(3\theta)\ d\theta \tag3$$
Is this setup correct?
There is problem in your "Area section of blue petal".
Imagine line $\theta =\dfrac{\pi}{12}$(join origin radially to intersection point) .Now what you're doing is subtracting area bounded between below this line and above yellow region only. Thus leaving behind small area bounded between above this line and below red petal.
correct set up should be
$$\text{required area }= \frac12\int^{\frac{\pi}{6}}_04\cos^2(3\theta)\ d\theta -\color{red} 2\times\frac12\int^{\frac{\pi}{12}}_04\sin^2(3\theta)\ d\theta $$
$=\dfrac{\pi}{6}-\dfrac{\pi-2}{6}=\dfrac{1}{3}$