A curve C in polar form $r=f(\theta)$ is parameterized by $C(\theta) = \langle f(\theta)\cos(\theta),f(\theta) \sin(\theta)\rangle$ because the x- and y-coordinates are given by $x = r \cos(\theta)$ and $y=r \sin(\theta)$. Evaluate $\int_C(x-y)^2ds$ where C is given by $r= 2 \cos(\theta)$ , $0\le \theta\le\frac{\pi}{2}$.

I'm not sure I'm understanding this question. If $r= 2 cos(\theta)$, then $C(\theta) = \langle2\cos^2(\theta),2\cos(\theta) \sin(\theta)\rangle$

Since $C'(\theta) = \langle -4\cos(\theta)\sin(\theta), 2\cos(2\theta)\rangle$, $||C'(\theta)|| = 2$. The integral is set up like this: $$2\int_0^{\frac{\pi}{2}}(2\cos^2(\theta)-2\cos(\theta)\sin(\theta))^2d\theta$$

This is not an easy integral to do, unless I am missing something. Any tips?

  • $\begingroup$ I'm not entirely sure about how you wrote down the problem, however I'm pretty sure that $||C'(\theta)|| \neq 2$ according to your $C'$. $\endgroup$ – Zubzub Jul 14 '16 at 8:47
  • $\begingroup$ The derivative of $2\cos^2(\theta)$ is not $-4\sin(\theta)$. It is $-4\cos(\theta)\sin(\theta)$ $\endgroup$ – dibdub Jul 14 '16 at 9:14
  • $\begingroup$ Damn you're right. Sorry I had a brain fart... ! $\endgroup$ – Zubzub Jul 14 '16 at 9:15
  • $\begingroup$ Don't worry about it! No problem $\endgroup$ – dibdub Jul 14 '16 at 9:16

$$I=\int_0^{\frac{\pi}{2}}(2\cos^2\theta-\sin2\theta)^2d\theta$$ $$I=4\int_0^{\frac{\pi}{2}}\cos^4\theta d\theta+\int_0^{\frac{\pi}{2}}\sin^22\theta\,d\theta-8\int_0^{\frac{\pi}{2}}\cos^3\theta\sin\theta \,d\theta$$ use $$\sin^22\theta=\frac{1-\cos4\theta}{2}$$ $$\int_{0}^{\frac{\pi}{2}}\sin^{2x-1}\theta\cos^{2y-1}\theta\,d\theta=\frac12\beta(x,y)$$ Indeed $$\int_0^{\frac{\pi}{2}}\cos^4\theta d\theta=\frac12\beta(\frac 12,\frac 52)$$ and $$\int_0^{\frac{\pi}{2}}\cos^3\theta\sin\theta \,d\theta=\frac12\beta( 1,2)$$

Other way $$\cos^3\theta \sin\theta=(1-\sin^2\theta) \cos\theta \sin \theta$$ set $u=\sin\theta$ and


  • $\begingroup$ Made an edit to the problem, the limits for $\theta$ are 0 and $\frac{\pi}{2}$. Sorry about that! $\endgroup$ – dibdub Jul 14 '16 at 8:23
  • $\begingroup$ It was edited... $\endgroup$ – Behrouz Maleki Jul 14 '16 at 8:25
  • $\begingroup$ I don't understand what you did there. Where do the last three equations come from? What does the β mean? I haven't learned that... $\endgroup$ – dibdub Jul 14 '16 at 8:27
  • $\begingroup$ $\beta(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ $\endgroup$ – Behrouz Maleki Jul 14 '16 at 8:29
  • $\begingroup$ I haven't learned any of this. What does that mean? $\endgroup$ – dibdub Jul 14 '16 at 8:30

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