Question

The polar curve $r = 1 + \cos \theta$ is rotated once around the point with polar coordinates $(2,0).$ What is the area of the region it sweeps over?

Answer 1

Answer:

$Area = -2.3147$

Step-by-step explanation:

Given

$r = 1 + \cos \theta$

Required

Determine the area with coordinates $(2,0)$

The area is represented as:

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

Where

$r = 1 + \cos \theta$

and

$(a,b) = (2,0)$

Substitute values for r, a and b in

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)^2} \, d\theta$

Expand

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)(1 + cos\theta)} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + 2cos\theta+cos^2\theta} )\, d\theta$

By integratin the above, we get:

$Area = \frac{1}{2}*\frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{2}[0,2]$

$Area = \frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{4}[0,2]$

Substitute 0 and 2 for $\theta$ one after the other

$Area = \frac{(cos(0) + 4)sin(0) + 3*0}{4} - \frac{(cos(2) + 4)sin(2) + 3*2}{4}$

$Area = \frac{(cos(0) + 4)sin(0)}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{(1 + 4)*0}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{-sin(2)(cos(2) + 4) - 6}{4}$

Get sin(2) and cos(2) in radians

$Area = \frac{-0.9093 * (-0.4161 + 4) - 6}{4}$

$Area = \frac{-9.2588}{4}$

$Area = -2.3147$