Question

Find the area of the region in the polar coordinate plane that is Inside the cardioid r = 2(1 + sin θ) and outside the circle r = 2sin θ

Answer 1

5π

Answer 2

The area of a region in polar coordinates is given by $A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta$. For the area between two polar curves $r_1(\theta)$ and $r_2(\theta)$, where $r_1$ is the outer curve and $r_2$ is the inner curve, the area is $A = \frac{1}{2} \int_{\alpha}^{\beta} (r_1^2 - r_2^2) d\theta$.

The two curves are:

1. Cardioid: $r_c(\theta) = 2(1 + \sin \theta)$
2. Circle: $r_s(\theta) = 2\sin \theta$

First, let's determine the relationship between the two curves. For the cardioid, $r_c(\theta) = 2 + 2\sin \theta$. Since $\sin \theta \ge -1$, $r_c(\theta) \ge 0$ for all $\theta$. The cardioid is traced for $\theta \in [0, 2\pi]$. For the circle, $r_s(\theta) = 2\sin \theta$. This curve is non-negative for $\theta \in [0, \pi]$ and negative for $\theta \in (\pi, 2\pi)$. The curve $r=2\sin\theta$ traces a circle of radius 1 centered at $(0,1)$ in Cartesian coordinates ($x^2 + (y-1)^2 = 1$).

We need to find the area inside the cardioid and outside the circle. This means we are looking for the area of points $(r, \theta)$ such that $0 \le r \le 2(1 + \sin \theta)$ and the point is not within the region enclosed by the circle $r = 2\sin \theta$.

Let's compare the radial distances of the two curves: $r_c(\theta) - r_s(\theta) = 2(1 + \sin \theta) - 2\sin \theta = 2 + 2\sin \theta - 2\sin \theta = 2$. Since $r_c(\theta) - r_s(\theta) = 2 > 0$, the cardioid is always radially further from the origin than the circle, i.e., $r_c(\theta) > r_s(\theta)$ for all $\theta$.

The circle $r = 2\sin \theta$ is entirely contained within the cardioid $r = 2(1 + \sin \theta)$. To see this, consider the area enclosed by each curve: Area of the cardioid $r = a(1 + \sin \theta)$ is $\frac{3}{2}\pi a^2$. For $a=2$, the area is $\frac{3}{2}\pi (2^2) = 6\pi$. Area of the circle $r = 2\sin \theta$ (which is a circle of radius 1) is $\pi (1)^2 = \pi$.

Since the circle is entirely inside the cardioid, the area of the region inside the cardioid and outside the circle is the difference between their total areas.

Area = Area of Cardioid - Area of Circle Area = $6\pi - \pi = 5\pi$.

Alternatively, using the integration formula: For $\theta \in [0, \pi]$, $\sin \theta \ge 0$, so both $r_c(\theta)$ and $r_s(\theta)$ are non-negative. In this interval, $r_c(\theta) > r_s(\theta)$. The area in this interval is: $A_1 = \frac{1}{2} \int_{0}^{\pi} [ (2(1 + \sin \theta))^2 - (2\sin \theta)^2 ] d\theta$ $A_1 = \frac{1}{2} \int_{0}^{\pi} [ 4(1 + 2\sin \theta + \sin^2 \theta) - 4\sin^2 \theta ] d\theta$ $A_1 = \frac{1}{2} \int_{0}^{\pi} (4 + 8\sin \theta) d\theta$ $A_1 = \int_{0}^{\pi} (2 + 4\sin \theta) d\theta$ $A_1 = [2\theta - 4\cos \theta]_{0}^{\pi} = (2\pi - 4\cos \pi) - (0 - 4\cos 0) = (2\pi - 4(-1)) - (-4) = 2\pi + 4 + 4 = 2\pi + 8$.

For $\theta \in (\pi, 2\pi)$, $\sin \theta < 0$, so $r_s(\theta) < 0$. The cardioid $r_c(\theta) = 2(1 + \sin \theta)$ is still non-negative. The region "outside the circle $r=2\sin\theta$" when $r_s(\theta)$ is negative means points not in the disk $x^2+(y-1)^2 \le 1$. Since the entire circle is contained within the cardioid, and the cardioid is always traced with $r \ge 0$, the region for $\theta \in (\pi, 2\pi)$ where the cardioid is defined and outside the disk is simply the area of the cardioid in this range.

However, the standard interpretation of "area inside curve A and outside curve B" when B is entirely inside A is Area(A) - Area(B). The integral $A_1 = 2\pi + 8$ represents the area between the two curves in the upper half-plane. The total area of the cardioid is $6\pi$, and the area of the circle is $\pi$. Since the circle is fully contained within the cardioid, the area inside the cardioid and outside the circle is $6\pi - \pi = 5\pi$.

The integral calculation for $A_1$ represents the area of the region where $2\sin\theta \le r \le 2(1+\sin\theta)$ for $\theta \in [0, \pi]$. This is only a part of the total area. The complete area is obtained by considering the entire region enclosed by the cardioid and subtracting the entire region enclosed by the circle, given that the circle is completely inside the cardioid.