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[tex] ds=\frac{ds}{d\theta}d\theta \mbox{ where } \frac{ds}{d\theta}=\sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} \\ [/tex]. Now [tex] \frac{dx}{d\theta}=-3a cos^2(\theta)sin(\theta) \\

\frac{dy}{d\theta}=3a sin^2(\theta)cos(\theta)[/tex]. Therefore [tex] \frac{ds}{d\theta}=\sqrt{9a^2(cos^4\theta sin^2\theta + sin^4\theta cos^2\theta)} \\ [/tex]. I simplify the expression as follows:

[tex] \frac{ds}{d\theta}=3a \sqrt{ cos^2\theta(\frac{1}{2}(1 + cos2\theta))(\frac{1}{2}(1 - cos2\theta)) + sin^2\theta(\frac{1}{2}(1 - cos2\theta))(\frac{1}{2}(1 + cos2\theta))} \\ [/tex]. This simplifies to

[tex] ds=3a\sqrt{(\frac{1}{4} - \frac{1}{4}cos^22\theta)(sin^2\theta + cos^2\theta)} d\theta[/tex]. Therefore the integral required is: [tex] \int_0^{\pi}2\pi6a^2\sqrt{\frac{1}{4} - \frac{1}{4}cos^22\theta}sin^3\theta d\theta[/tex]

(1)Am I on the correct path as regards simplifying the express I got for [tex]ds/d\theta [/tex]?

(2) Did I simplify the expression correctly? Because the integral = zero with me.