Find the volume V obtained by rotating the region bounded by the curves y = 3 sin x, y = 3 cos x, 0 ≤ x ≤ (pi/4) about y = 3

We are asked to find the volume \( V \) obtained by rotating the region bounded by the curves \( y = 3 \sin(x) \) and \( y = 3 \cos(x) \), for \( 0 \leq x \leq \frac{\pi}{4} \), about the line \( y = 3 \).

### Step-by-step process:

1. **Identify the region:**
   The region is bounded between the curves \( y = 3 \sin(x) \) and \( y = 3 \cos(x) \), and between \( x = 0 \) and \( x = \frac{\pi}{4} \).

2. **Determine the method of rotation:**
   Since we are rotating about \( y = 3 \), which is parallel to the x-axis, we will use the washer method. The washer method is applicable when the region is rotated around a horizontal axis, and the volume is given by:

   \[
   V = \pi \int_{a}^{b} \left[ R_{\text{outer}}(x)^2 - R_{\text{inner}}(x)^2 \right] dx
   \]

   where \( R_{\text{outer}}(x) \) is the distance from the axis of rotation to the outer curve and \( R_{\text{inner}}(x) \) is the distance to the inner curve.

3. **Set up the radii:**
   - The outer curve is \( y = 3 \cos(x) \).
   - The inner curve is \( y = 3 \sin(x) \).
   - The axis of rotation is \( y = 3 \).

   The radii are computed by finding the vertical distances from the axis of rotation to the curves:
   - \( R_{\text{outer}}(x) = 3 - 3 \cos(x) \) (distance from the axis to \( y = 3 \cos(x) \)).
   - \( R_{\text{inner}}(x) = 3 - 3 \sin(x) \) (distance from the axis to \( y = 3 \sin(x) \)).

4. **Set up the volume integral:**

   The volume is given by:
   \[
   V = \pi \int_0^{\frac{\pi}{4}} \left[ (3 - 3 \sin(x))^2 - (3 - 3 \cos(x))^2 \right] dx
   \]

5. **Simplify and solve the integral:**

   Let's expand the squares inside the integral:

   \[
   (3 - 3 \sin(x))^2 = 9(1 - 2 \sin(x) + \sin^2(x))
   \]
   \[
   (3 - 3 \cos(x))^2 = 9(1 - 2 \cos(x) + \cos^2(x))
   \]

   Substituting these into the volume formula:

   \[
   V = \pi \int_0^{\frac{\pi}{4}} \left[ 9(1 - 2 \sin(x) + \sin^2(x)) - 9(1 - 2 \cos(x) + \cos^2(x)) \right] dx
   \]

   Simplifying the integrand:

   \[
   V = \pi \int_0^{\frac{\pi}{4}} 9 \left[ -2 \sin(x) + \sin^2(x) + 2 \cos(x) - \cos^2(x) \right] dx
   \]

   Factor out the 9:

   \[
   V = 9\pi \int_0^{\frac{\pi}{4}} \left[ -2 \sin(x) + \sin^2(x) + 2 \cos(x) - \cos^2(x) \right] dx
   \]

6. **Separate the integrals:**

   Break the integral into four separate parts:

   \[
   V = 9\pi \left( \int_0^{\frac{\pi}{4}} -2 \sin(x) \, dx + \int_0^{\frac{\pi}{4}} 2 \cos(x) \, dx + \int_0^{\frac{\pi}{4}} \sin^2(x) \, dx - \int_0^{\frac{\pi}{4}} \cos^2(x) \, dx \right)
   \]

7. **Evaluate each integral:**

   - \( \int_0^{\frac{\pi}{4}} \sin(x) \, dx = 1 - \frac{1}{\sqrt{2}} \)
   - \( \int_0^{\frac{\pi}{4}} \cos^

This problem involves finding the volume of a solid obtained by rotating the region bounded by y = 3 sin(x) and y = 3 cos(x), from 0 to π/4, about the line y = 3. The solution uses the washer method, covering concepts of trigonometry and calculus. Suitable for Grades 11-12 or early college students.

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