객관식
How many sides does a polygon have if the sum of its interior angles is ?
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7. Non-Right Triangles
Area of SAS & ASA Triangles
객관식
Given the polar curve , what is the area enclosed by one loop of the curve?
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검증된 단계별 안내1
Recall that the area enclosed by one loop of a polar curve given by \(r = f(\theta)\) is calculated using the formula:
\(\text{Area} = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta\)
where \(\alpha\) and \(\beta\) are the limits of integration corresponding to one loop of the curve.
Identify the function and the limits for one loop. Here, \(r = 4 \sin(5\theta)\). Since the sine function completes one full period over \(2\pi\), and the argument is \(5\theta\), one loop corresponds to the interval where \(\sin(5\theta)\) goes from 0 to its next zero, i.e., from \(\theta = 0\) to \(\theta = \frac{\pi}{5}\).
Set up the integral for the area of one loop:
\(\text{Area} = \frac{1}{2} \int_0^{\frac{\pi}{5}} (4 \sin(5\theta))^2 \, d\theta = \frac{1}{2} \int_0^{\frac{\pi}{5}} 16 \sin^2(5\theta) \, d\theta\).
Use the trigonometric identity to simplify the integrand:
\(\sin^2 x = \frac{1 - \cos(2x)}{2}\), so
\(16 \sin^2(5\theta) = 16 \times \frac{1 - \cos(10\theta)}{2} = 8 (1 - \cos(10\theta))\).
Rewrite the integral and integrate term-by-term:
\(\text{Area} = \frac{1}{2} \int_0^{\frac{\pi}{5}} 8 (1 - \cos(10\theta)) \, d\theta = 4 \int_0^{\frac{\pi}{5}} (1 - \cos(10\theta)) \, d\theta\).
Then compute the integral of \(1\) and \(\cos(10\theta)\) over the interval, and simplify to find the area enclosed by one loop.
