Area of the region bounded by $r = \sin \theta$
1,721
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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Comments

phquiett 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} (12\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:
$\frac{\pi}{2}$
$\frac{\pi}{4}$
$\pi$
$1$
$2$
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 yearsThe 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.

phquiett almost 4 years@leastaction even putting the first line in symbolab gives me $\frac{\pi}{4}$

Ninad Munshi almost 4 yearsSomehow, 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 $2\pi$.