Find the area of the region cut from the first quadrant by the curve requals6 left parenthesis 3 minus sine 2 theta right parenthesis Superscript one half
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To find the area of the region cut from the first quadrant by the polar curve \( r = 6 \left( 3 - \sin(2\theta) \right)^{\frac{1}{2}} \), we can use the formula for the area of a region in polar coordinates:

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

### Step-by-Step Solution:

1. **Square the equation for \(r\):**
   \[
   r^2 = 6^2 \left( 3 - \sin(2\theta) \right) = 36(3 - \sin(2\theta)) = 108 - 36\sin(2\theta)
   \]

2. **Set the limits of integration:**
   Since the region is in the first quadrant, the limits for \( \theta \) range from 0 to \( \frac{\pi}{2} \). This is because the first quadrant spans from \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \) in polar coordinates.

3. **Set up the area integral:**
   \[
   A = \frac{1}{2} \int_0^{\frac{\pi}{2}} \left( 108 - 36\sin(2\theta) \right) \, d\theta
   \]

4. **Evaluate the integral:**

   First, break the integral into two parts:
   \[
   A = \frac{1}{2} \left[ \int_0^{\frac{\pi}{2}} 108 \, d\theta - \int_0^{\frac{\pi}{2}} 36\sin(2\theta) \, d\theta \right]
   \]

   - The first part:
     \[
     \int_0^{\frac{\pi}{2}} 108 \, d\theta = 108 \left[ \theta \right]_0^{\frac{\pi}{2}} = 108 \times \frac{\pi}{2} = 54\pi
     \]

   - The second part:
     \[
     \int_0^{\frac{\pi}{2}} 36\sin(2\theta) \, d\theta = 36 \times \left[ -\frac{\cos(2\theta)}{2} \right]_0^{\frac{\pi}{2}} = 36 \times \left[ -\frac{\cos(\pi)}{2} + \frac{\cos(0)}{2} \right] 
     \]
     \[
     = 36 \times \left[ -\frac{-1}{2} + \frac{1}{2} \right] = 36 \times 0 = 0
     \]

5. **Final area calculation:**
   \[
   A = \frac{1}{2} \times 54\pi = 27\pi
   \]

Thus, the area of the region in the first quadrant is \( 27\pi \) square units.

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Would you like more details or clarifications on any step?

Here are 5 related questions to explore:

1. How do you find the area for a region in polar coordinates in general?
2. What happens to the area if the curve includes both positive and negative sine terms?
3. How would the limits change if the region were in a different quadrant?
4. What is the significance of symmetry in polar areas, and how does it affect the integral?
5. Can you use this method to find the area of any general polar curve?

**Tip:** When solving area problems in polar coordinates, always remember to check the curve's symmetry to simplify calculations.

Find the area of the region cut from the first quadrant by the curve r = 6(3 - sin(2θ))^1/2.

Learn how to find the area of a region in the first quadrant bounded by the polar curve r = 6(3 - sin(2θ))^1/2. This detailed solution explains how to use integration in polar coordinates to calculate the area, covering key concepts like limits of integration and simplifying the integral. Suitable for high school students in Grades 11-12.

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