Let $D$ be an region under cycloid (which are you are finding). So, you should compute an integral:
$$\iint_{D} 1 \; dx dy$$
In wikipedia we can find Green's theorem, so if we choose for example $M(x,y)=2x$ and $L(x,y)=y$ we have:
$$\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}=2-1=1$$
So:
$$\iint_{D}\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\;dxdy=\iint 1\; dx dy$$
By Green's theorem:
$$\iint_{D}\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\;dxdy=\oint_{C} M \; dx+ L\; dy$$
where $M,L$ are like above. $C$ is curve around $D$. Let's find form of $C$: there are two parts: segment and cycloid. Now we should find two points where cycloid touch $y$ axis, that means solution of:
$$y=0=1-\cos(t)$$
so $t=0$ and $t=2\pi$. Putting value of $t$ into second equation we have:
$$x=t-\sin t=0-0=0$$
and
$$x=2\pi-\sin(2\pi)=2\pi$$
So: $C=[0,2\pi] \cup $ arc of cycloid. So it is sufficient to compute integral:
$$\oint_{C} M \; dx+ L\; dy$$
along $C$. Can you finish this solution?
EDIT: Original answer didn't address radial function taking on negative values.
The positive direction for the radial polar variable $r$ is outward so if you want the area of the region outside the polar curve $r = f(\theta) \geq 0$ and inside the curve $r = g(\theta)$, i.e.
$$
f(\theta) \leq r \leq g(\theta)
$$
for all $\alpha \leq \theta \leq \beta$, then you calculate the definite integral
$$
A = \frac12 \int_\alpha^\beta \bigl[ g(\theta)^2 - f(\theta)^2 \bigr] \, \mathrm{d}\theta.
$$
Let $f(\theta) = 3\cos\theta$ and $g(\theta) = 1 + \cos\theta$.
Both functions exhibit the symmetry $f(2\pi - \theta) = f(\theta)$ and $g(2\pi - \theta) = g(\theta)$ for all $\theta$, which is equivalent to the fact that their polar graphs have a reflection symmetry across $\theta = \pi$, the $x$-axis.
Thus, we can calculate the total area for
$\frac\pi3 \leq \theta \leq \frac{5\pi}3$
by calculating the area for
$\frac\pi3 \leq \theta \leq \pi$
and doubling the result.
However, there's still an issue: $f(\theta) \leq 0$ for $\frac\pi2 \leq \theta \leq \pi$, so the points wrap around the bottom half of the circle. The naive application of the integral formula gives negative values over this domain. See this picture:
Instead, we have to calculate the integral in two pieces:
\begin{array}{ccr}
\tfrac\pi3 \leq{} \theta \leq \tfrac\pi2
& \quad\leadsto & f(\theta) \leq{} r \leq g(\theta) \\
\tfrac\pi2 \leq{} \theta \leq \pi
& \quad\leadsto & 0 \leq{} r \leq g(\theta)
\end{array}
Thus, the total area is
\begin{align}
A &= 2 \cdot \frac12 \biggl( \int_{\pi/3}^{\pi/2}
\bigl[ g(\theta)^2 - f(\theta)^2 \bigr] \, \mathrm{d}\theta
\;+\; \int_{\pi/2}^{\pi}
g(\theta)^2 \, \mathrm{d}\theta \biggr) \\
&= \int_{\pi/3}^{\pi/2}
\bigl[ (1 + \cos\theta)^2 - (3\cos\theta)^2 \bigr] \, \mathrm{d}\theta
\;+\; \int_{\pi/2}^{\pi}
(3\cos\theta)^2 \, \mathrm{d}\theta
\end{align}
You can probably take it from here.
Best Answer
I think you want
If you input $a=0,b=\pi/2$, using the above property you can "convert" sines to cosines and vice versa due to $\sin x=\cos (\pi/2 -x)$.
But what you have done, as pointed out by others, is not applicable everywhere. If you ''exchange" sines and cosines using the above property, it is totally fine.