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$Find the area A of the shaded region of the cardioid r=7-7cos(theta). The graph shows the graph of the cardioid with a small region shaded. The x and y axes are unlabelled. (Express numbers in exact form. Use symbolic notation and fractions where needed.)$

Find the area $A$ of the shaded region of the cardioid $r=7-7 \cos (\theta)$ . $\newline$ The graph shows the graph of the cardioid with a small region shaded. The $\mathrm{x}$ and $\mathrm{y}$ axes are unlabelled. $\newline$ (Express numbers in exact form. Use symbolic notation and fractions where needed.)

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Find equations of tangent lines using limits

Full solution

Q. Find the area

A

of the shaded region of the cardioid

r=7-7 \cos (\theta)

.

\newline

The graph shows the graph of the cardioid with a small region shaded. The

\mathrm{x}

and

\mathrm{y}

axes are unlabelled.

\newline

(Express numbers in exact form. Use symbolic notation and fractions where needed.)

Set Up Integral: To find the area of the shaded region of the cardioid, we first need to set up the integral for the area of one loop of the cardioid. The formula for the area $A$ enclosed by a polar curve $r(\theta)$ from $\theta = a$ to $\theta = b$ is: $\newline$ $A = \frac{1}{2} \int_{a}^{b} r(\theta)^2 \, d\theta$ $\newline$ For the cardioid $r = 7 - 7\cos(\theta)$ , we need to integrate from $\theta = 0$ to $\theta = 2\pi$ to cover the entire curve.
Substitute and Expand: Substitute $r = 7 - 7\cos(\theta)$ into the area formula: $\newline$ $A = \frac{1}{2} \int_{0}^{2\pi} (7 - 7\cos(\theta))^2 \, d\theta$ $\newline$ Expanding the square: $\newline$ $(7 - 7\cos(\theta))^2 = 49 - 98\cos(\theta) + 49\cos^2(\theta)$ $\newline$ So, the integral becomes: $\newline$ $A = \frac{1}{2} \int_{0}^{2\pi} (49 - 98\cos(\theta) + 49\cos^2(\theta)) \, d\theta$
Evaluate Integral: Evaluate the integral: $\newline$ The integral of a constant over a period is just the constant times the period, and the integral of $\cos(\theta)$ over a full period is zero. The integral of $\cos^2(\theta)$ over a full period is $\pi$ . Therefore: $\newline$ $\int_{0}^{2\pi} 49 \, d\theta = 98\pi$ $\newline$ $\int_{0}^{2\pi} -98\cos(\theta) \, d\theta = 0$ $\newline$ $\int_{0}^{2\pi} 49\cos^2(\theta) \, d\theta = 49\pi$ $\newline$ Adding these up: $\newline$ $A = \frac{1}{2} (98\pi + 0 + 49\pi) = \frac{1}{2} \times 147\pi = 73.5\pi$

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