Rose (mathematics) - Area

Area

A rose whose polar equation is of the form

where k is a positive integer, has area

\frac{1}{2}\int_{0}^{2\pi}(a\cos (k\theta))^2\,d\theta = \frac {a^2}{2} \left(\pi + \frac{\sin(4k\pi)}{4k}\right) = \frac{\pi a^2}{2}

if k is even, and

\frac{1}{2}\int_{0}^{\pi}(a\cos (k\theta))^2\,d\theta = \frac {a^2}{2} \left(\frac{\pi}{2} + \frac{\sin(2k\pi)}{4k}\right) = \frac{\pi a^2}{4}

if k is odd.

The same applies to roses with polar equations of the form

since the graphs of these are just rigid rotations of the roses defined using the cosine.

Read more about this topic:  Rose (mathematics)

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