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please express as a double integraj\l the area enclosed by one loop of the curve r=3cos2\theta
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\(r=3\cos2\theta\) or \(r=3\cos^2\theta\)?
r=3\cos2\theta
Here's a sketch of one loop: |dw:1409680881343:dw| This region can be defined by the set, \[R:=\left\{(r,\theta)~:~-\frac{\pi}{4}\le\theta\le\frac{\pi}{4},~0\le r\le 3\cos2\theta\right\}\] As a double integral, the area is then \[\large A=\int_{-\pi/4}^{\pi/4}\int_0^{3\cos2\theta}~r~dr~d\theta\] Using symmetry, you can evaluate it as \[\large A=2\int_0^{\pi/4}\int_0^{3\cos2\theta}~r~dr~d\theta\]
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