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

Question

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?

Answer 1

Note the area integral is

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

Answer 2

The Cartesian equation for $r=\sin\theta$ is given by

$$x^2 + \left(y-\frac{1}{2}\right)^2 = \frac{1}{4}$$

however the Cartesian equation for $r=|\sin\theta|$ is given by two circles, not one, of the form

$$x^2 + \left(y\pm\frac{1}{2}\right)^2 = \frac{1}{4}$$

Thus the area bounded by this polar curve in the $xy$-plane is twice the area of one circle, $2\cdot\left(\frac{\pi}{4}\right) = \frac{\pi}{2}$

Answer 3

Check out this Desmos graph to see which step in your integration process is wrong: https://www.desmos.com/calculator/9wklwenvqf