Problem 42 - Calculus Early Transcendentals

Question

Sketch each region and use a double integral to find its area. The region inside both the cardioid \(r=1+\sin \theta\) and the cardioid \(r=1+\cos \theta\)

Step-by-Step Solution

Answer

Question: Find the area of the region formed by the intersection of the cardioids given by the polar equations \(r=1+\sin(\theta)\) and \(r=1+\cos(\theta)\). Answer: The area of the region formed by the intersection of the two cardioids is \(A = 2 - \frac{1}{2}\pi\) square units.

Step 1: Sketch the Cardioids

First, let's sketch the cardioids given by the polar equations \(r=1+\sin(\theta)\) and \(r=1+\cos(\theta)\). You can use graphing software or a graphing calculator to plot these curves. The sketch should look like: [](https://postimg.cc/Hj5Kzxfr) We need to find the area of the region formed by the intersection of these cardioids.

Step 2: Find the Points of Intersection

To find the points of intersection between the two cardioids, we need to solve the system of equations given by: \(r_1 = 1 + \sin(\theta)\) \(r_2 = 1 + \cos(\theta)\) We should find the common values of \(\theta\) where both equations are equal. Solve the equations, \(1 + \sin(\theta) = 1 + \cos(\theta)\), which simplifies to, \(\sin(\theta) = \cos(\theta)\). The solutions to this equation are \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{5\pi}{4}\).

Step 3: Set up the Double Integral

We will use a double integral to find the area of the region formed by the intersection of the cardioids. The area can be expressed as the integrals: \(A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}\int_{1+\sin(\theta)}^{1+\cos(\theta)}r\;dr\;d\theta\)

Step 4: Solve the Double Integral

Now, we will solve the double integral: \(A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \left[\frac{1}{2}r^2\Bigg\rvert_{1+\sin(\theta)}^{1+\cos(\theta)}\right] d\theta\) \(= \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{1}{2}[(1+\cos(\theta))^2 - (1+\sin(\theta))^2]\;d\theta\) This integral is a bit tricky to solve, and you might need to use some trigonometric identities to simplify it. After simplification, we get: \(A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \left[2\cos(\theta)\sin(\theta) + \cos^2(\theta) - \sin^2(\theta) \right] d\theta\) Now integrate over \(\theta\): \(A = 2 - \frac{1}{2}\pi\) Thus, the area of the region formed by the intersection of the two cardioids is \(A = 2 - \frac{1}{2}\pi\) square units.