# How do you find the area ( if any ) common to the four cardioids r=1+-cos theta and r = 1+-sin theta?

Sep 29, 2016

$\frac{3}{2} \pi - 4 \sqrt{2} + 1 = 0.055535$ areal units.

#### Explanation:

The lines of symmetry (axes ) of the cardioids are the positive and

negative axes of coordinates.

The cardioids are equal in size and symmetrically placed, with

respect to the pole r=0 that is a common point of imtersection for all.

The common area comprises four equal parts.

The other terminal points, in these parts, are

$\left(1 - \frac{1}{\sqrt{2}} , \frac{\pi}{4}\right) \in {Q}_{1}$

$\left(1 - \frac{1}{\sqrt{2}} , \frac{3}{4} \pi\right) \in {Q}_{2}$

$\left(1 - \frac{1}{\sqrt{2}} , \frac{5}{4} \pi\right) \in {Q}_{3}$

$\left(1 - \frac{1}{\sqrt{2}} , \frac{7}{4} \pi\right) \in {Q}_{4}$

Now, one such part is the area in ${Q}_{3}$ that is

symmetrical about $\theta = \frac{5}{4} \pi$ and bounded by

$r = 1 + \cos \theta$, between $\left(0 , \pi\right) \mathmr{and} \left(1 - \frac{1}{\sqrt{2}} , \frac{5}{4} \pi\right)$ and

$r = 1 + \sin \theta$, between $\left(0 , \frac{3}{2} \pi\right) \mathmr{and} \left(1 - \frac{1}{\sqrt{2}} , \frac{5}{4} \pi\right)$

This area = $2 \int \int r \mathrm{dr} d \theta$,

r from 0 to $\left(1 + \cos \theta\right) \mathmr{and} \theta$ from $\pi$ t.o $\frac{5}{4} \pi$.

After integration with respect to r, this becomes

$\int {\left(1 + \cos \theta\right)}^{2} d \theta$, between the limits $\pi \mathmr{and} \frac{5}{4} \pi .$

$= \int \left(1 + 2 \cos \theta + {\cos}^{2} \theta\right) d \theta$, between the limits

$= \int \left(1 + 2 \cos \theta + \frac{1 + \cos 2 \theta}{2}\right) d \theta$, between the limits

$= \left[\frac{3}{2} \theta + 2 \sin \theta + \frac{1}{4} \sin 2 \theta\right]$, between $\pi \mathmr{and} \frac{5}{4} \pi .$

$= \frac{3}{8} \pi - \frac{2}{\sqrt{2}} + \frac{1}{4}$

$= \frac{3 \pi - 8 \sqrt{2} + 2}{8}$

The total common area is 4 X this area

$= \frac{3}{2} \pi - 4 \sqrt{2} + 1 = 0.055535$ areal units

I welcome a graphical depiction for my answer..

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