Questions: Use a double integral to find the area of the region bounded by the cardioid (r=5(1-cos theta)). Set up the double integral as efficiently as possible, in polar coordinates, that is used to find the area. (r drdtheta ) (Type exact answers, using (pi) as needed.)

Solution Steps

To find the area of the region bounded by the cardioid \( r = 5(1 - \cos \theta) \), we can set up a double integral in polar coordinates. The area \( A \) can be found by integrating the function \( r \) over the region defined by the cardioid. The limits for \( r \) will be from 0 to \( 5(1 - \cos \theta) \), and the limits for \( \theta \) will be from 0 to \( 2\pi \).

1. Set up the double integral in polar coordinates to find the area.
2. Integrate with respect to \( r \) from 0 to \( 5(1 - \cos \theta) \).
3. Integrate with respect to \( \theta \) from 0 to \( 2\pi \).

The cardioid is given by the polar equation \( r = 5(1 - \cos \theta) \). This describes a heart-shaped curve in polar coordinates.

To find the area \( A \) enclosed by the cardioid, we set up the double integral in polar coordinates: \[ A = \int_0^{2\pi} \int_0^{5(1 - \cos \theta)} r \, dr \, d\theta \]

First, we evaluate the inner integral: \[ \int_0^{5(1 - \cos \theta)} r \, dr = \left[ \frac{r^2}{2} \right]_0^{5(1 - \cos \theta)} = \frac{(5(1 - \cos \theta))^2}{2} = \frac{25(1 - \cos \theta)^2}{2} \]

Next, we evaluate the outer integral: \[ A = \int_0^{2\pi} \frac{25(1 - \cos \theta)^2}{2} \, d\theta \] This simplifies to: \[ A = \frac{25}{2} \int_0^{2\pi} (1 - 2\cos \theta + \cos^2 \theta) \, d\theta \] Using the identity \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \), we can further simplify: \[ A = \frac{25}{2} \left( \int_0^{2\pi} 1 \, d\theta - 2\int_0^{2\pi} \cos \theta \, d\theta + \frac{1}{2} \int_0^{2\pi} (1 + \cos(2\theta)) \, d\theta \right) \] Calculating these integrals gives: \[ \int_0^{2\pi} 1 \, d\theta = 2\pi, \quad \int_0^{2\pi} \cos \theta \, d\theta = 0, \quad \int_0^{2\pi} \cos(2\theta) \, d\theta = 0 \] Thus, we have: \[ A = \frac{25}{2} \left( 2\pi + 0 + \frac{1}{2}(2\pi) \right) = \frac{25}{2} \left( 2\pi + \pi \right) = \frac{25}{2} \cdot 3\pi = \frac{75\pi}{2} \]

Final Answer