Area of the region bounded by $r = |\sin \theta|$


Note the area integral is

$$A=\int_0^{2\pi} \int_0^{|\sin(\theta)|} rdrd\theta =\frac12 \int_0^{2\pi} \sin^2\theta d\theta =\frac14 \int_0^{2\pi} (1-\cos2\theta )d\theta=\frac\pi2$$


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Updated on January 27, 2023


  • ph-quiett
    ph-quiett 9 months

    I'm trying to find the area of the region bounded by $r =|\sin \theta | $ in the $xy$-plane using the formula: \begin{align*} A &= \frac{1}{2}\int_{0}^{\pi} (|\sin \theta|)^2 d\theta \\ &=\frac{1}{2}\int_{0}^{\pi} (1-\cos2\theta)^2 d\theta \\ &= \frac{1}{2}\int_{0}^{\pi} (1-2\cos\theta + \cos^22\theta)^2 d\theta \\ \end{align*}

    which I get $\frac{\pi}{4}$ as the answer but I got it wrong on a test.

    The other options are:

    1. $\frac{\pi}{2}$

    2. $\frac{\pi}{4}$

    3. $\pi$

    4. $1$

    5. $2$

    Could there have been a mistake on the test or did I miss something?

    • leastaction
      leastaction almost 4 years
      $1-\cos(2\theta) = 2 \sin^2\theta$.
    • Ninad Munshi
      Ninad Munshi almost 4 years
      The other problem is that your integral does not get you the full area. $r=\sin\theta$ is a circle of radius $\frac{1}{2}$, but $r=|\sin\theta|$ is two circles.
    • ph-quiett
      ph-quiett almost 4 years
      @leastaction even putting the first line in symbolab gives me $\frac{\pi}{4}$
    • Ninad Munshi
      Ninad Munshi almost 4 years
      Somehow, your incorrect manipulation got the correct result for the integral, an example of two mistakes canceling each other out. Even still, the limits of integration are wrong, it should be from $0$ to $2\pi$.