Mathematics, 13.07.2019 23:30 emj70
Find the area of the region enclosed by one loop of the curve. r = 3/2 sin(2θ)
Answers: 1
which corresponds to one loop.![A=\frac{9}{4}* \frac{1}{2} \int_0^ {\pi/2} sin^2 2\theta d\theta\\ \\ A=\frac{9}{8}\int_0^ {\pi/2} \frac{1-cos 4\theta}{2} d \theta\\ \\ A=\frac{9}{8} [\theta-\frac{sin 4\theta}{2} ]_0^{\pi/2}\\ \\ A=\frac{9}{8}*\pi/2\\ \\ A=\frac{9 \pi}{16}\\](/tpl/images/0086/5511/bc28b.png)
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Find the area of the region enclosed by one loop of the curve. r = 3/2 sin(2θ)...
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