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What is the area of the region in the plane bounded by the curve given in polar coordinates $r = 4 + 2\cos(2\theta)$?

Could someone walk me through the conversion of polar coordinates to rectangular coordinates or how to integrate when just given a polar equation?

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  • $\begingroup$ For polar functions, we are usually calculating the area enclosed by the curve (or some portion of it), rather than the area between the curve and a coordinate axis, or between two curves as with Cartesian coordinates. So a double integral is usually unnecessary. $\endgroup$ – colormegone Mar 28 '14 at 6:04
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You have the heuristic (and graphic) result:

the part of the graphic in color is close to a triangle with both sides of length $f(\theta)$ and angle $\delta\theta$;

its area is $$\frac 12 f(\theta)^2 \sin\delta\theta\simeq \frac 12 f(\theta)^2 \delta\theta $$ Actually the exact formula is $$ A = \frac 12\int_{\theta_1}^{\theta_2} f(\theta)^2 d\theta. $$

A proper justification would be, from the knowledge of the Jacobian of the polar transformation: $$ A = \int_{\theta_1}^{\theta_2} d\theta \int_{0}^{f(\theta)} rdr = \int_{\theta_1}^{\theta_2} d\theta \frac {f(\theta)^2}{2} = \frac 12 \int_{\theta_1}^{\theta_2} f(\theta)^2 d\theta. $$


In your case, $$A = \frac 12 \int_0^{2\pi} (4+2\cos2\theta)^2 d\theta = 2\pi \frac{16 + 0 + 2}2 = 18\pi. $$

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