객관식
How many sides does a polygon have if the sum of its interior angles is ?
149
views
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?
A
B
C
D
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.
관련 영상
관련 실천
객관식
Given the polar curves and , what is the area of the region that lies inside both curves?
113
views
객관식
Given the polar curves and , what is the area of the region that lies inside both curves?
106
views
