How do I find the exact length of the polar curve $$r = 1+sin(\theta)$$ from $$\frac{\pi}{3} \leq \theta \leq \pi $$?

I had originally setup my equation as $$\int_{\frac{\pi}{3}}^{\pi} \sqrt{(1+\sin(\theta) )^2 + (\cos\theta)^2} * d\theta$$ but that got me the incorrect answer. What am I doing wrong?

  • 1
    $\begingroup$ Are you sure of the definition of the length of a polar curve? $\endgroup$ – Mhenni Benghorbal Apr 11 '16 at 3:07
  • $\begingroup$ @MhenniBenghorbal Yes $\endgroup$ – Greencat Apr 11 '16 at 3:22
  • $\begingroup$ Your expression looks right, but the parentheses are hard to read. We want to integrate the square root of $(1+\sin\theta)^2+\cos^2\theta$, so we want to integrate $\sqrt{2+2\sin\theta}$. $\endgroup$ – André Nicolas Apr 11 '16 at 3:24
  • $\begingroup$ @AndréNicolas sorry. I fixed the parenthesis. Could you explain that a bit more? $\endgroup$ – Greencat Apr 11 '16 at 3:28
  • $\begingroup$ @imranfat that is the derivative of r (necessary for the equation being used) $\endgroup$ – Greencat Apr 11 '16 at 3:28

As I mentioned in comments, we want $$\int_{\pi/3}^{\pi}\sqrt{2+2\sin\theta}\,d\theta.$$ Let $\theta=\frac{\pi}{2}-t$. Then $\sin\theta=\cos t$, and after the substitution we want $$\int_{-\pi/2}^{\pi/6}\sqrt{2+2\cos t}\,dt.$$ But by the identity $\cos(2y)=2\cos^2y-1$, we have $2+2\cos t=4\cos^2(t/2)$. So we want $$\int_{-\pi/2}^{\pi/6}2\cos(t/2)\,dt.$$ This integral is straightforward. Note that the $\sin(\pi/12)$ we get as a component of the answer can be expressed exactly in terms of square roots, so the final answer has a nice simple form.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.