How do you find the area of $r=1+cos(theta)$ ?

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Answer 1 · 2016-07-23T03:14:41

$(3pi)/2$ areal units.

If the pole r = 0 is not outside the region, the area is given by

$(1/2) int r^2 d theta$ , with appropriate limits.

The given curve is a closed curve called cardioid.

It passes through the pole r = 0 and is symmetrical about the initial

line $theta = 0$ .

As $r = f(cos theta)$ , r is periodic with period $2pi$ .

And so the area enclosed by the cardioid is

$(1/2) int r^2 d theta$ , over $theta in [0, 2pi]$ .

$(1/2)(2) int (1+cos theta)^2 d theta, theta in [0, pi]$ , using symmetry about $theta=0$

$=int (1+cos theta)^2 d theta, theta in [0, pi]$

$=int (1+2 cos theta + cos^2theta) d theta, theta in [0, pi]$

$=int (1+2 cos theta + (1+cos 2theta)/2) d theta, theta in [0, pi]$

$=[3/2theta+2sin theta]+(1/2)(1/2)sin 2theta]$ ,

between limits $0 and pi$

$=(3pi)/2+0+0$

How do you find the area of r=1+cos(theta)r=1+cos(theta)?