Find the area of the following regions given in polar coordinates ($r$=radius, $t$=angle) A) $D={(r;t) : 1 + cos(t) \le r \le 3cos(t)}$ B) $D={(r;t) : r \le |2cos(5t)|}$
In A) I solved the inequality $1 + cos(t) \le 3cos(t)$ keeping in mind that $r$ must be bigger than 0 and $t \epsilon [0;2 \pi ]$ and got that $t \epsilon [0; \frac{ \pi}{3}]U[ \frac{5 \pi }{3};2 \pi]$. Then I found the area of $3cos(t)$ between 0 and $\frac{ \pi}{3}$ (which I called $Ar_2$), and between $\frac{5 \pi }{3}$ and $2 \pi$ (which I called $Ar_2')$. After that, I did the same thing but with $r= 1 + cos(t)$, and I called the areas $Ar_1$ and $Ar_1'$ respectively.
Then, I computed $(Ar_2 - Ar_1)+(Ar_2' -Ar_1)$ to get the total area the one that I had to calculate. Is it ok?
In B) what should I do? Should I solve $-2cos(5t) \le 2cos(5t)$ and compute the same that in A?
Notice that the area that you are interested in looks like:
Were the lines represent the values $t$ for which $r=0$: for instance $t=\pi/10$ and $t=3\pi/10$ in the first quadrant. Notice that, in the first quadrant:
$$|2cos 5t|= \begin{cases} 2cos 5t &\text{ if} & t \in[0,\pi/10]\cup[3\pi/10,\pi/2]\\ -2cos 5t &\text{ if} & t \in[\pi/10,3\pi/10] \end{cases} $$
This means that we can compute the desired area as follows:
$$A=\int\int_D r\, dr\,dt$$
Given our drawing, for the first quadrant this is equal to:
\begin{align} A&=\int_0^{\pi/10}\int_0^{2cos 5t} r\, dr\,dt+\int_{\pi/10}^{3\pi/10}\int_0^{-2cos 5t} r\, dr\,dt+\int_{3\pi/10}^{\pi/2}\int_0^{2cos 5t} r\, dr\,dt\\ &=\frac{1}{2}\int_0^{\pi/10}(2cos 5t)^2 dt+\frac{1}{2}\int_{\pi/10}^{3\pi/10}(-2cos 5t)^2 dt+\frac{1}{2}\int_{3\pi/10}^{\pi/2}(2cos 5t)^2 dt\\ &=\pi/2 \end{align}
This means that the total area given by the four quadrants is given by $4\cdot A=2\pi$. Similarly notice that each petal has area $\pi/5$ and there are 10 petals.