Multiple Choice
Which of the following integrals correctly represents the area of the region enclosed by the curves , , and for ?
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9. Graphical Applications of Integrals
Area Between Curves
Multiple Choice
Find the area enclosed by one loop of the curve .
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D
Verified step by step guidance1
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 = \(\sin\)(6\(\theta\)). To find the bounds of integration (\(\theta\)_1 and \(\theta\)_2), determine the values of \(\theta\) where one loop of the curve is completed. A loop occurs when r returns to zero after completing a full cycle. Solve \(\sin\)(6\(\theta\)) = 0 to find these bounds.
Step 3: Solve \(\sin\)(6\(\theta\)) = 0. The solutions are \(\theta\) = 0, \(\frac{\pi}{6}\), \(\frac{\pi}{3}\), etc. For one loop, the bounds are \(\theta\)_1 = 0 and \(\theta\)_2 = \(\frac{\pi}{6}\). These bounds will be used in the integral.
Step 4: Substitute r = \(\sin\)(6\(\theta\)) into the area formula. The integral becomes A = \(\frac{1}{2}\) \(\int\)_{0}^{\(\frac{\pi}{6}\)} (\(\sin\)(6\(\theta\)))^2 \, d\(\theta\). To simplify the integral, use the trigonometric identity (\(\sin\)(x))^2 = \(\frac{1 - \cos(2x)}{2}\).
Step 5: Rewrite the integral using the identity: A = \(\frac{1}{2}\) \(\int\)_{0}^{\(\frac{\pi}{6}\)} \(\frac{1 - \cos(12\theta)}{2}\) \, d\(\theta\). Simplify further and evaluate the integral step by step to find the area enclosed by one loop of the curve.
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