I need help finding the length of the curve represented by this particular relation.

Question

I need help finding the length of the curve represented by the following relation: $$x = 5\,cos^3\theta; y = 5\,sin^3 \theta$$

Here is what I've tried: $$s = \int_0^{2\pi} \sqrt{(\frac{d\theta}{d\theta})^2 + (\frac{d\theta}{d\theta})^2}\,d\theta$$ $$\frac{d\theta}{d\theta} = -15\,sin\theta\,cos^2\,\theta$$ $$\frac{dy}{d\theta} = 15\,sin^2\theta\,cos\,\theta$$

Plugging in,

$$s = \int_0^{2\pi} \sqrt{(-15\,sin\theta\,cos^2\,\theta)^2 + (15\,sin^2\theta\,cos\,\theta)^2}\,d\theta$$ $$s = \int_0^{2\pi} \sqrt{225\,sin^2\,\theta\,cos^4\,\theta + 225\,sin^4\,\theta\,cos^2\,\theta}\,d\theta$$ $$s = \int_0^{2\pi} \sqrt{225\,sin^2\,\theta\,cos^2\,\theta\,(cos^2\,\theta + sin^2\,\theta)}\,d\theta$$ $$s = \int_0^{2\pi} \sqrt{225\,sin^2\,\theta\,cos^2\,\theta}\,d\theta$$ $$s = \int_0^{2\pi} 15\,sin\,\theta\,cos\,\theta\,d\theta$$ $$s = 15\,\int_0^{2\pi} \,sin\,\theta\,cos\,\theta\,d\theta$$

And there I am stumped... The textbook says that the answer should be 30, and my graphing application corroborates that number. I know I can't get to 30 from here. What am I doing wrong?

Answer 1

Have a look to the graph of the function $$\sqrt{225\,\sin^2(\theta)\,\cos^2(\theta)}$$ over the range $0 \leq \theta\leq 2\pi$. It is composed of four arches. Then, do what kobe suggested.

Answer 2

Your mistake was in the second to the last step where your integrand is $15\sin \theta \cos \theta$. It should be $15|\sin \theta\cos \theta|$. You'll need to split up your integral over the intervals $[0,\pi/2]$, $[\pi/2, \pi]$, $[\pi, 3\pi/2]$, and $[3\pi/2, 2\pi]$.