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

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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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### ph-quiett

hotter than july in California

Updated on January 27, 2023

• 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 almost 4 years
$1-\cos(2\theta) = 2 \sin^2\theta$.
• 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 almost 4 years
@leastaction even putting the first line in symbolab gives me $\frac{\pi}{4}$
• 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 $2\pi$.