Finding bounds of integration non-graphically I have this curve given in polar form:
$$
\rho=4\sin2\phi
$$
I'm trying to find the integration bounds of the area it creates, like this:
$$
t = 2\phi \\
t \in [0 + 2k\pi, \pi+2k\pi], k\in Z \\
0+2k\pi \le t \le\pi+2k\pi \\
0+2k\pi \le 2\phi \le\pi+2k\pi \\
k\pi \le \phi \le \frac{\pi}{2}+k\pi \\
0 \le \phi \le \frac{\pi}{2} \vee \pi \le \phi \le \frac{3\pi}{2}\\
$$
The base period of sin function is $2\pi$ and $\sin \ge 0$ only at $\phi \in [0 + 2k\pi, \pi+2k\pi]$, which means that $4\sin2\phi \ge 0$ when $\phi \in [k\pi, \frac{\pi}{2}+k\pi]$. So that's why I've written the last line as my integration bounds, but now I know that those are not all of them. Can someone explain it to me?
 A: No need to consider all $k,$ because you want area of a single petal of the rose and it is possible to consider symmetry for congruent areas. Radius $\rho $ vanishes when $ \sin 2 \phi=0\rightarrow 2\phi= 0, \pi. $ This will be clear while finding tangents when $\rho=0$
Polar angles $\theta$ at origin set as limits of tangential positions :
$$\rho= a \sin 2 \theta\tag1$$
Angles $(\psi,\theta)$  are expressed by relations
$$ \rho'= \cos \psi, \theta'= \dfrac{\sin \psi }{\rho} \tag2$$
Differentiating on arc length using above differential relations
$$  \cos \psi = \rho'= 2a \cos 2 \theta  \cdot \theta' \tag3$$
Simplify
$$\tan \psi = \dfrac12 \tan  2 \theta \tag4 $$
Tangent at origin occurs in this case at
$$  (\psi= 0,\; \phi=0,\; \rho=0 \;;) \tag5 $$
$$ 2 \theta = \tan ^{-1} 0= 0, \pi \tag6$$
So tangency at origin occurs at:
$$ \theta= (0,\pi/2) \tag7$$
Total Area
$=8\int_0^{\pi/4} \frac12 \rho^2 d\phi $ Plug in $\rho $ to evaluate Area further.
To check slope of tangent at any point $\gamma= \theta+\psi$
