8. Consider the polar curve given by the equation \( r=2 \sin (\theta) \) a) (7pts) The point \( \left(\sqrt{2}, \frac{\pi}{4}\right) \) is a point on this polar curve. Determine the \( x \) )-coordinates of this point. b) (8pts) Represent this curve as a parametric curve (with parameter \( \theta \) ). c) (8pts) Set up, but do not evaluate, an integral which represents the area enclosed by this curve.

To begin, the polar curve \( r = 2 \sin(\theta) \) describes a circle centered at \( (0, 1) \) with a radius of 1 in Cartesian coordinates. The point \(\left(\sqrt{2}, \frac{\pi}{4}\right)\) can be converted using \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \), leading us to find the \( x \)-coordinate. For part b, the parametric representation can be obtained from \( r = 2 \sin(\theta) \). In Cartesian coordinates, using \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \), we find that the parametric equations are \( x(\theta) = 2 \sin(\theta) \cos(\theta) \) and \( y(\theta) = 2 \sin^2(\theta) \). For part c, the area \( A \) enclosed by one loop of the polar curve can be calculated using the formula \( A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta \). Thus, we set up the integral as \( A = \frac{1}{2} \int_{0}^{\pi} (2 \sin(\theta))^2 \, d\theta \), where \( \alpha = 0 \) and \( \beta = \pi \). --- Did you know that polar coordinates come in handy when dealing with circular shapes? This system simplifies matters, letting us express curves like \( r = 2 \sin(\theta) \) with ease. Understanding these coordinates unlocks a treasure chest of functions related to angles and distance from the origin, making calculations smoother and geometrically intuitive. If you find yourself intrigued by curves like this, diving into "Calculus" by James Stewart or any college-level precalculus book can deepen your knowledge. Online resources like video lectures or websites dedicated to math can also provide engaging ways to learn, ensuring you grasp the beauty of functions beyond just equations!