$3$ . Find the area in the interior of $r=1-\sin \theta$ above the polar axis. Use your calculator to integrate.

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3

. Find the area in the interior of

r=1-\sin \theta

above the polar axis. Use your calculator to integrate.

Set up integral limits: To find the area inside the curve $r = 1 - \sin(\theta)$ above the polar axis, we need to set up an integral with the appropriate limits of integration. The curve crosses the polar axis at $\theta = \frac{\pi}{2}$ and $\theta = \frac{3\pi}{2}$ . Since we are looking for the area above the polar axis, we will integrate from $\theta = 0$ to $\theta = \pi$ .
Area formula for polar curve: The formula for the area $A$ enclosed by a polar curve $r(\theta)$ from $\theta = a$ to $\theta = b$ is given by $A = \frac{1}{2} \int_{a}^{b} (r(\theta))^2 d\theta$ . For our curve $r = 1 - \sin(\theta)$ , the area $A$ above the polar axis is $A = \frac{1}{2} \int_{0}^{\pi} (1 - \sin(\theta))^2 d\theta$ .
Expand integrand: We need to expand the integrand $(1 - \sin(\theta))^2$ before integrating. Expanding gives us $(1 - \sin(\theta))^2 = 1 - 2\sin(\theta) + \sin^2(\theta)$ . Now we can integrate each term separately.
Integrate $1$ and $-2\sin(\theta)$ : The integral of $1$ with respect to $\theta$ from $0$ to $\pi$ is simply $\theta$ evaluated from $0$ to $\pi$ , which gives us $\pi$ . The integral of $-2\sin(\theta)$ with respect to $\theta$ from $0$ to $\pi$ is $1$ $3$ evaluated from $0$ to $\pi$ , which gives us $1$ $6$ . The integral of $1$ $7$ is more complex and requires a trigonometric identity.
Integrate $\sin^2(\theta)$ : We use the power-reduction identity $\sin^2(\theta) = \frac{1}{2} - \frac{1}{2}\cos(2\theta)$ to integrate $\sin^2(\theta)$ . The integral of $\sin^2(\theta)$ with respect to $\theta$ from $0$ to $\pi$ is then the integral of $\frac{1}{2} - \frac{1}{2}\cos(2\theta)$ from $0$ to $\pi$ , which is $\sin^2(\theta) = \frac{1}{2} - \frac{1}{2}\cos(2\theta)$ $0$ evaluated from $0$ to $\pi$ .
Evaluate $\sin^2(\theta)$ integral: Evaluating the integral of $\frac{1}{2} - \frac{1}{2}\cos(2\theta)$ from $0$ to $\pi$ gives us $\frac{1}{2}\pi - \frac{1}{4}\sin(2\pi) + \frac{1}{4}\sin(0) = \frac{1}{2}\pi$ . The $\sin$ terms vanish because $\sin(2\pi) = \sin(0) = 0$ .
Sum up integrals: Now we add up the results of the integrals: $\pi$ (from the integral of $1$ ) - $4$ (from the integral of $-2\sin(\theta)$ ) + $\frac{1}{2}\pi$ (from the integral of $\sin^2(\theta)$ ). This gives us $\pi - 4 + \frac{1}{2}\pi = \frac{3}{2}\pi - 4$ .
Final area calculation: Finally, we multiply the result by $1/2$ to get the area $A$ , because the formula for the area in polar coordinates includes a $1/2$ factor. So, $A = 1/2 \times (3/2\pi - 4) = 3/4\pi - 2$ .