Calculate the area bounded by the polar curve $ r(\theta)=2+\sin(\theta)$

Question

How can I calculate the area bounded by the following curve:

$$ r(\theta)=2+\sin(\theta) $$

Answer 1

HINT...You need to calculate using the standard formula for polar areas, namely $$\frac 12\int r^2 d\theta$$

Therefore in this case you need to evaluate $$\frac 12\int_0^{2\pi}(2+\sin\theta)^2 d\theta$$

Answer 2

If we call your region $A$ the area is given by $$ \iint_A dxdy. $$ To change variables from Cartesian coordinates to polar coordinates we have to add the Jacobian of this transformation such that $dxdy = r dr d\theta$. The integral can be expressed as $$ \int_{0}^{2\pi} \int_{0}^{2 + \sin(\theta)} r dr d\theta. $$ Now it is just integrating $$ \int_{0}^{2\pi} \int_{0}^{2 + \sin(\theta)} r dr d\theta= \int_{0}^{2\pi}\left[\frac{\sin ^2(\theta)}{2}+2 \sin (\theta)+2\right] d\theta=\frac{9}{2}\pi. $$ This answer is believable if we look at the following plot.

Here the red line gives a circle of radius $2$ around the point $(0,1)$. This circle has an area of $4\pi$. We would expect your area, whose boundary is given by the blue line, to be a little bit bigger. So $\frac{9}{2}\pi = (4+\frac{1}{2})\pi$ makes sense.