analysisareacalculuscurvesintegration
I have to generalize the concept of finding the region $D$ inside of limaçons without a loop. In particular $r=a+b\cos(\theta)$, where $0\leq\theta\leq2\pi$, and to prevent the loop $a>b>0$.
My naive approach is to find: $$\int\int_{D}1dxdy = \int_{a-b}^{a+b}\int_{0}^{2\pi}rd\theta dr$$ But then I loose information that $r$ is $a+b\cos\theta$.
I checked how the limaçons with different $a$ and $b$ look like via desmos, but this does not give me any intuition.
Draw a picture. We use the standard formula for area in polar coordinates. In principle there could be a problem with the interval of integration, but here there is no problem, because $3+\sin\theta$ is always positive. So our area is $$\int_0^{2\pi}\frac{1}{2}(3+\sin\theta)^2\,d\theta.$$ For the details, expand. The integral of $\frac{9}{2}$ is easy. For the "middle" term, the integral is easily $0$. Finally, we need to deal with $\int_0^{2\pi}\frac{1}{2}\sin^2\theta\,d\theta$. There are many ways to do this. One of them is to use $\cos2\theta=1-2\sin^2\theta$. Another way is to use symmetry, and note that the integral over our interval of $\cos^2\theta$ is the same as the integral of $\sin^2\theta$. But the sum of these integrals is the integral of $1$.
Here is a polar plot of the function $r(\theta) =7+14 \cos \theta$ for $\theta \in [0, 2 \pi)$.
Note that the curve passes through the origin when $r(\theta) = 0$. Solving this for $\theta \in [0, 2 \pi)$ gives solutions $\pi \pm { \pi \over 3}$.
Hence you need to compute $A = {1 \over 2}\int_{2 \pi \over 3}^{4 \pi \over 3} r^2(\theta) d \theta$.
Scroll over for answer:
Computing gives $A = {7^2 \over 2} (2 \pi -3 \sqrt{3})$.
Best Answer
By analogy to the rectangular integral to find the area of $\int y\; dx$ the usual way to find the area in polar coordinates is $\int_0^{2 \pi} \frac 12r^2\; d\theta$. The area element is approximately a triangle with height $r$ and base $r\; d\theta$, so area $\frac 12 r^2 d\theta$. Now you substitute in your formula for $r$ to get $\frac 12\int_0^{2\pi} (a+b\cos \theta)^2\; d\theta$. A good picture is here.