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Multiple Choice
Find the area of the region enclosed by one loop of the curve .
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Step 1: Recognize that the problem involves finding the area enclosed by one loop of a polar curve. The formula for the area enclosed by a polar curve is A = \(\frac{1}{2}\) \(\int\)_{\(\theta\)_1}^{\(\theta\)_2} r^2 \, d\(\theta\), where r is the polar function and \(\theta\)_1, \(\theta\)_2 are the bounds of integration.
Step 2: Identify the polar function r = \(\sin\)(8\(\theta\)). To find the bounds of integration, determine the values of \(\theta\) where one loop of the curve is completed. For r = \(\sin\)(8\(\theta\)), a loop is completed when \(\sin\)(8\(\theta\)) returns to zero after one full oscillation. Solve \(\sin\)(8\(\theta\)) = 0 to find \(\theta\)_1 and \(\theta\)_2.
Step 3: Substitute r = \(\sin\)(8\(\theta\)) into the area formula. The integral becomes A = \(\frac{1}{2}\) \(\int\)_{\(\theta\)_1}^{\(\theta\)_2} (\(\sin\)(8\(\theta\)))^2 \, d\(\theta\). Use the trigonometric identity (\(\sin\)(x))^2 = \(\frac{1 - \cos(2x)}{2}\) to simplify the integrand.
Step 4: Simplify the integral further. The integral now becomes A = \(\frac{1}{2}\) \(\int\)_{\(\theta\)_1}^{\(\theta\)_2} \(\frac{1 - \cos(16\theta)}{2}\) \, d\(\theta\). Break this into two separate integrals: A = \(\frac{1}{4}\) \(\int\)_{\(\theta\)_1}^{\(\theta\)_2} 1 \, d\(\theta\) - \(\frac{1}{4}\) \(\int\)_{\(\theta\)_1}^{\(\theta\)_2} \(\cos\)(16\(\theta\)) \, d\(\theta\).
Step 5: Evaluate each integral. The first integral \(\int\)_{\(\theta\)_1}^{\(\theta\)_2} 1 \, d\(\theta\) gives the length of the interval \(\theta\)_2 - \(\theta\)_1. The second integral \(\int\)_{\(\theta\)_1}^{\(\theta\)_2} \(\cos\)(16\(\theta\)) \, d\(\theta\) can be solved using the formula \(\int\) \(\cos\)(ax) \, dx = \(\frac{\sin(ax)}{a}\). Combine the results to find the area A.