Integrate curve given in polar coords solution verification My solution to the below problem is $4\pi$ but the answer sheet says it's $8\pi$. Please verify my calculations:
$$
\rho=4\sin2\phi \implies 4\sin2\phi \ge0 \implies \sin t\ge 0
$$
Where $t = 2\phi$
Now, I'm trying to find the integration bounds like this:
$$
t \in [0 + 2k\pi, \pi+2k\pi], k\in Z \\
0+2k\pi \le t \le\pi+2k\pi \\
0+2k\pi \le 2\phi \le\pi+2k\pi \\
k\pi \le \phi \le \frac{\pi}{2}+k\pi \\
0 \le \phi \le \frac{\pi}{2} \vee \pi \le \phi \le \frac{3\pi}{2}\\
$$
Therefore the integrals should be:
$$
\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 16\sin^22\phi\  d\phi + \frac{1}{2} \int_{\pi}^{\frac{3\pi}{2}} 16\sin^22\phi\  d\phi = \\
8 \int_{0}^{\frac{\pi}{2}} \sin^22\phi\ d\phi + 8 \int_{\pi}^{\frac{3\pi}{2}} \sin^22\phi\  d\phi
$$
The indefinite integral is:
$$
\int \sin^22\phi\ d\phi = \frac{1}{2}\phi - \frac{1}{8}\sin4\phi + C
$$
And so I continue the calculations:
$$
8\left[\frac{1}{2}\phi-\frac{1}{8}\sin4\phi\right]_{0}^{\frac{\pi}{2}} + 8\left[\frac{1}{2}\phi-\frac{1}{8}\sin4\phi\right]_{\pi}^{\frac{3\pi}{2}} = 4\pi
$$
This got to be something simple...
EDIT
The problem I'm trying to solve is to calculate the area of a figure bounded by a curve given by the first equation in this question.

This is the bound -
i) For $0 \le \phi \le \frac{\pi}{2}, 0 \le 2\phi \le \pi$.
ii) For $\frac{\pi}{2} \le \phi \le \pi, \pi \le 2\phi \le 2\pi$.
iii) For $\pi \le \phi \le \frac{3\pi}{2}, 2\pi \le 2\phi \le 3\pi$.
iv) For $\frac{3\pi}{2} \le \phi \le 2\pi, 3\pi \le 2\phi \le 4\pi$.
This is rose curve with $4$ petals.
(i) gives you the petal in first quadrant, (ii) gives you the petal in the fourth quadrant and so on...
Given we have to find area bound by this curve, let's find area of one of the petals and multiply by $4$.
Area of the petal in the first quadrant $= \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 16\sin^22\phi\  d\phi $
$= 8 \int_{0}^{\frac{\pi}{2}} \sin^22\phi\  d\phi = 8\left[\frac{1}{2}\phi-\frac{1}{8}\sin4\phi\right]_{0}^{\frac{\pi}{2}} = 2\pi$
Total area $= 4 \times 2\pi = 8\pi$
Or you can apply the bound (ii), (iii), (iv) in the integral to get to total area $8\pi$.
