91Ó°ÊÓ

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.

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

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

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.