Find the area of the polar region enclosed by f(θ)=sin(θ) for 0≤θ≤π/6


To find the area of a region enclosed by a polar curve, \( r = f(\theta) \), over a given range of \( \theta \), we use the following formula for the area in polar coordinates:

\[
A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta
\]

Given \( f(\theta) = \sin(\theta) \) and the range \( 0 \leq \theta \leq \frac{\pi}{6} \), the formula becomes:

\[
A = \frac{1}{2} \int_{0}^{\frac{\pi}{6}} \sin^2(\theta) \, d\theta
\]

### Step 1: Simplify the integral
We can use the identity:

\[
\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}
\]

Now the integral becomes:

\[
A = \frac{1}{2} \int_{0}^{\frac{\pi}{6}} \frac{1 - \cos(2\theta)}{2} \, d\theta
\]

Simplifying further:

\[
A = \frac{1}{4} \int_{0}^{\frac{\pi}{6}} (1 - \cos(2\theta)) \, d\theta
\]

### Step 2: Integrate the terms
Now, we can integrate each term:

\[
A = \frac{1}{4} \left[ \theta - \frac{1}{2} \sin(2\theta) \right]_{0}^{\frac{\pi}{6}}
\]

### Step 3: Evaluate the integral
Now we evaluate the definite integral:

\[
A = \frac{1}{4} \left[ \frac{\pi}{6} - \frac{1}{2} \sin\left( \frac{\pi}{3} \right) - \left( 0 - \frac{1}{2} \sin(0) \right) \right]
\]

\[
A = \frac{1}{4} \left( \frac{\pi}{6} - \frac{1}{2} \times \frac{\sqrt{3}}{2} \right)
\]

\[
A = \frac{1}{4} \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)
\]

### Step 4: Final expression for the area
\[
A = \frac{\pi}{24} - \frac{\sqrt{3}}{16}
\]

This is the area of the polar region enclosed by \( f(\theta) = \sin(\theta) \) for \( 0 \leq \theta \leq \frac{\pi}{6} \).

Would you like any further details on the steps or explanation?

Here are some related questions:
1. How do you find the area between two polar curves?
2. How does symmetry affect area calculation in polar coordinates?
3. What other trigonometric identities simplify polar integrals?
4. Can polar integrals always be evaluated with elementary functions?
5. How does the area of this region change if the angle is increased?

**Tip:** Always check if a polar curve exhibits symmetry—it simplifies integration!

Find the area of the polar region enclosed by f(θ)=sin(θ) for 0≤θ≤π/6

Learn how to calculate the area of a polar region enclosed by the curve f(θ)=sin(θ) for the range 0≤θ≤π/6. This solution involves integrating in polar coordinates and applying trigonometric identities. Suitable for advanced high school students or college-level calculus learners.

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