Find the area of the region lying inside the polar curve r = 2(1 − cos theta) and outside the polar curve r + 4 cos(theta)= 0.

To find the area of the region, we need to calculate the difference between the areas enclosed by the two polar curves.

First, let's find the intersection points of the curves r = 2(1 − cos(theta)) and r + 4 cos(theta) = 0.

Substituting r = 2(1 − cos(theta)) into the equation r + 4 cos(theta) = 0, we get:

2(1 − cos(theta)) + 4 cos(theta) = 0

Expanding and simplifying:

2 − 2cos(theta) + 4cos(theta) = 0
2cos(theta) = 2
cos(theta) = 1
theta = 0 (since cos(0) = 1)

So, the polar curves intersect at theta = 0.

To find the area inside the curve r = 2(1 − cos(theta)) and outside the curve r + 4 cos(theta) = 0, we need to evaluate the integral of (1/2)(r^2 d(theta)) over the interval where 0 ≤ theta ≤ 2π. However, since the curves only intersect at theta = 0, we need to split the integral into two parts:

Part 1: Evaluate ∫(1/2)(r^2 d(theta)) for the region between 0 ≤ theta ≤ 2π, excluding the segment from theta = 0 to the first intersection point.

Part 2: Evaluate ∫(1/2)(r^2 d(theta)) for the region between 0 ≤ theta ≤ 2π, excluding the segment from the second intersection point to theta = 2π.

For Part 1, we can integrate as follows:

∫ (1/2)(r^2 d(theta)) = ∫ (1/2)(2(1 − cos(theta))^2 d(theta))
= ∫ (1/2)(4(1 − 2cos(theta) + cos^2(theta)) d(theta))
= ∫ (2 − 4cos(theta) + 2cos^2(theta)) d(theta)
= ∫ 2 d(theta) − ∫ 4cos(theta) d(theta) + ∫ 2cos^2(theta) d(theta)

The first integral is simply 2(theta). The second integral is -4sin(theta), and the third integral is 2(theta/2 + sin(2theta)/4). Thus, for Part 1, the integrated result is:

2(theta) − 4sin(theta) + (theta/2 + sin(2theta)/4)

Evaluating this expression from theta = 0 to the first intersection point, we get:

[2(theta) − 4sin(theta) + (theta/2 + sin(2theta)/4)] evaluated from 0 to the first intersection point

= [2(0) − 4sin(0) + (0/2 + sin(2(0))/4)] − [2(0) − 4sin(0) + (0/2 + sin(2(0))/4)]

= 0

For Part 2, we can integrate as follows:

∫ (1/2)(r^2 d(theta)) = ∫ (1/2)(2(1 − cos(theta))^2 d(theta))

Following the same procedure as Part 1, the integrated result for Part 2 is:

2(theta) − 4sin(theta) + (theta/2 + sin(2theta)/4)

Evaluating this expression from the second intersection point to theta = 2π, we get:

[2(theta) − 4sin(theta) + (theta/2 + sin(2theta)/4)] evaluated from the second intersection point to 2π

= [2(2π) − 4sin(2π) + (2π/2 + sin(2(2π))/4)] − [2(theta) − 4sin(theta) + (theta/2 + sin(2theta)/4)] evaluated from the first intersection point to 2π

= [4π − 4sin(2π) + (2π/2 + sin(2(2π))/4)] − [0]

= 4π + π = 5π

Therefore, the area of the region inside the curve r = 2(1 − cos(theta)) and outside the curve r + 4 cos(theta) = 0 is 5π.

To find the area of the region lying inside the polar curve r = 2(1 - cos θ) and outside the polar curve r + 4cos(θ) = 0, we need to find the points of intersection between these two curves.

First, let's find the points of intersection by setting the two equations equal to each other:

2(1 - cos θ) = -4cos(θ)

Now, we can solve this equation to find the values of θ at which the curves intersect. However, since this equation is not easy to solve algebraically, we can use a numerical method called the bisection method or a graphing calculator to find the approximate values of θ.

Once we have the values of θ at the points of intersection, we can find the corresponding values of r by substituting the values of θ into either of the original equations.

Now, we have the polar coordinates of the points of intersection. To find the area between the curves, we need to integrate the difference between the two curves over the range of angles where they intersect.

The formula for finding the area between two polar curves is:

A = ∫[θ1, θ2] (1/2)([r2^2 - r1^2]) dθ

where r1 and r2 are the two curves.

Using this formula, we can integrate the difference between the two curves over the range of θ from the first point of intersection to the second point of intersection to find the area.

Depending on the specific values, the integration process can be quite tedious. So, evaluating the integral using numerical methods or a computer algebra system is recommended.