User:Alexbrewer/CalculusSandbox

1973BC40
The formula used to determine the area of a region enclosed by a polar curve is: $$A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta \;; \qquad 0<\beta-\alpha\le2\pi$$

Apply the formula to the equation given. This curve is a cardioid, so the interval is $$\left ( 0,2\pi\right ]$$

$$A = \frac{1}{2} \int_{0}^{2\pi} (1-cos(\theta))^2 d\theta$$

Square the contents of the parentheses:

$$= \frac{1}{2} \int_{0}^{2\pi} 1-2cos(\theta)+cos^2(\theta)\; d\theta$$

Integrate with respect to $$\theta$$:

$$= \frac{1}{2} \left [ \frac{sin(x)cos(x)}{2}-2sin(x)+ \frac{3x}{2} \right ] \bigg| \begin{matrix} 2\pi \\ 0 \end{matrix}$$

Evaluate on the interval $$\left ( 0,2\pi\right ]$$:

$$= \frac{1}{2} \;3\pi \quad = \frac{3\pi}{2} \quad = \quad (C) \quad \frac{3}{2}\pi$$