How do I compute a enclosed area using double integral with polar coordinates?

Question

Say I am given two functions $f(r,\theta)$ and $g(r,\theta)$.

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Let $f(r,\theta)$ be the red function and $g(r,\theta)$ be the blue function. Therefore, $f(r,\theta)>g(r,\theta)$. The points of intersection are: $(r_1,\theta_1)$ and $(r_2,\theta_2)$. How do I set up a double integral to calculate the enclosed area?

Answer 1

Hint:

You can split the area in two parts by a line of constant $\theta$, and integrate separately.

$$\iint_{f(\theta,\rho)<0,\\\theta_0<\theta<\theta_1}\rho\,d\rho\,d\theta.$$

Answer 2

I do not know of any implicit polar coordinate relations like

$$ \sin \theta/r + e^{ r \theta} =\frac12 \tag1 $$

(or in cartesian coordinates $\frac{y}{x^2+y^2}+e^{\sqrt{x^2+y^2} \tan^{-1}(y/x)}=\frac12$)

that is implied in your question. Neither does any CAS support it as of usual.

However, if $r_1=f(\theta_1),r_2=f(\theta_2)$ are separately available explicitly and their intersection points are $(a,\alpha)(b,\beta)(c,\gamma),$ then the common intersection area $A$ using single independent variable $\theta$ sweep out is:

$$2A=\int_\alpha^\beta r_1^2\ d\theta +\int_\beta^\gamma r_2^2\ d\theta \tag2 $$

In such a situation using double integral is of no particular advantage and so I am avoiding its use in this answer.

Double integral evaluation and implicit polar coordinates do not go together as happens with cartesian coordinates.

In this situation intersection point between curves can be solved

$$ r= (1+\sin \theta),\,r= 3(1-\sin \theta), \beta= \pi/6$$ and area

$$ 2A=\int_{-\pi/2}^{\pi/6} (1+\sin \theta)^2\ d\theta +\int_{\pi/6}^{\pi/2}3^2 (1-\sin\theta)^2\ d\theta \tag3 $$

which you can take further.

Answer 3

A curve representing 1) and cardioid $ r=(1-\sin \theta)$ is shown:

How to find the area between explicit and implicit polar (blue and red respectively) curves?