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 \(r=1+\cos \theta\)

Short 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-by-step solution

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.

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 \(\theta = \frac{5\pi}{4}\).

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: \(A = \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}\int_{1+\sin(\theta)}^{1+\cos(\theta)}r\;dr\;d\theta\)

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.