Question

Which integral yields the arc length of \(r=3(1-\cos 2 \theta)\) ? State why the other integrals are incorrect. (a) \(3 \int_{0}^{2 \pi} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\) (b) \(12 \int_{0}^{\pi / 4} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\) (c) \(3 \int_{0}^{\pi} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\) (d) \(6 \int_{0}^{\pi / 2} \sqrt{(1-\cos 2 \theta)^{2}+4 \sin ^{2} 2 \theta} d \theta\)

Short answer

The correct integral is thus (c). The others are incorrect since they either have wrong limits showing that they do not cover the entire curve (options a, b, and d) or they have the incorrect coefficient outside the integral (options a, b, and d).

Step-by-step solution

Analyze the given equation

The given polar equation is \(r=3(1-\cos 2 \theta)\). Notice that as \(\theta\) varies from 0 to \(2\pi\), the curve goes around twice, this is due to the \(2\theta\) term. Therefore, the limits of the integral should be from 0 to \(\pi\), and if necessary, multiplied by 2. This analysis should point to options (c) and (d).

Find the derivative of the given equation

Differentiating \(r = 3(1-\cos 2 \theta)\) with respect to \(\theta\) gives \(\frac {dr}{d\theta} = 6sin2\theta\).

Analyze the given integrals and rule out incorrect options

Looking at the integrals in the choices given, you can see that the term under the square root in each integral matches the terms of the standard formula mentioned earlier after plugging in \(r\) and \(\frac {dr}{d\theta}\). Thus, the integral would look like this: \(\sqrt {(3(1 - cos2\theta))^2 + (6sin2\theta)^2}\). The important part is to examine the coefficient outside the integral and the limits. The only integral to correctly includes '3' as the coefficient and the limits from 0 to \(\pi\) is option (c).