Question

Consider the three-petaled polar rose defined by $r=\cos 3\theta$.Explain why the definite integral localid="1653739667075" $\frac{1}{2}{\int }_0^{2\pi }{\cos }^{2}3\theta d\theta$calculates twice the area bounded by the petals of this rose.

Short answer

If the area is calculated in the interval $[0,2\pi ]$t gives twice the area bounded by the petals of the polar curve $r=\cos 3\theta$

Step-by-step solution

Given information

The definite integral is $\frac{1}{2}{\int }_0^{2\pi }{\cos }^{2}3\theta d\theta$

Consider the polar curve$r=\cos 3\theta$

The objective is to give the reason why the area of the curve 12∫02πcos23θdθ   is twice the actual area

The curve $r=\cos 3\theta$traced twice in the interval $[0,2\pi ]$

As a result, calculating the area in the interval$\left[0,2\pi \right]$

It delivers twice the area enclosed by the polar curve's$r=\cos 3\theta$ petals.