Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y = 7

y = sin(x) y = 0 0 <= x <= pi/2

pi(14 - pi/2)

○ pi(28 - pi/4)

○ pi(7 - pi/2)

○ pi(7 - pi/4)

pi(14 - pi/4)

We are tasked with finding the volume of the solid generated by revolving the region bounded by the equations \( y = \sin(x) \), \( y = 0 \), and \( 0 \leq x \leq \frac{\pi}{2} \), about the line \( y = 7 \).

### Step-by-Step Solution:

1. **Identify the Method**: This is a problem of finding the volume of a solid of revolution using the **washer method**. Since we are revolving around the line \( y = 7 \) (a horizontal line), the washer method is appropriate.

2. **Washer Method Formula**: The volume of a solid of revolution generated by revolving a region around a horizontal line \( y = c \) is given by:

   \[
   V = \pi \int_{a}^{b} \left[ R(x)^2 - r(x)^2 \right] dx
   \]

   where:
   - \( R(x) \) is the outer radius (distance from \( y = 7 \) to \( y = 0 \)),
   - \( r(x) \) is the inner radius (distance from \( y = 7 \) to \( y = \sin(x) \)),
   - \( a = 0 \), \( b = \frac{\pi}{2} \) (the limits of integration).

3. **Outer Radius**: The outer radius is constant, as the lower bound is \( y = 0 \) and the axis of rotation is \( y = 7 \):

   \[
   R(x) = 7 - 0 = 7
   \]

4. **Inner Radius**: The inner radius varies with \( x \), as it is the distance between \( y = 7 \) and \( y = \sin(x) \):

   \[
   r(x) = 7 - \sin(x)
   \]

5. **Set up the Volume Integral**:

   \[
   V = \pi \int_{0}^{\frac{\pi}{2}} \left[ 7^2 - (7 - \sin(x))^2 \right] dx
   \]

   Simplifying the integrand:
   \[
   7^2 = 49
   \]
   and
   \[
   (7 - \sin(x))^2 = 49 - 14\sin(x) + \sin^2(x)
   \]
   So, the integrand becomes:
   \[
   49 - (49 - 14\sin(x) + \sin^2(x)) = 14\sin(x) - \sin^2(x)
   \]

6. **Integrate**:

   \[
   V = \pi \int_{0}^{\frac{\pi}{2}} \left( 14\sin(x) - \sin^2(x) \right) dx
   \]

   Split the integral into two parts:
   \[
   V = \pi \left( \int_{0}^{\frac{\pi}{2}} 14\sin(x) dx - \int_{0}^{\frac{\pi}{2}} \sin^2(x) dx \right)
   \]

   - For \( \int_{0}^{\frac{\pi}{2}} 14\sin(x) dx \):
     \[
     \int_{0}^{\frac{\pi}{2}} 14\sin(x) dx = 14 \left[ -\cos(x) \right]_{0}^{\frac{\pi}{2}} = 14(0 - (-1)) = 14
     \]

   - For \( \int_{0}^{\frac{\pi}{2}} \sin^2(x) dx \), use the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \):
     \[
     \int_{0}^{\frac{\pi}{2}} \sin^2(x) dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \cos(2x)) dx
     \]
     \[
     = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}
     \]

7. **Final Calculation**:

   Now substitute these results into the volume equation:
   \[
   V = \pi \left( 14 - \frac{\pi}{4} \right)
   \]

Thus, the volume of the solid is:

\[
\boxed{\pi \left( 14 - \frac{\pi}{4} \right)}
\]

This corresponds to the answer: \( \pi(14 - \frac{\pi}{4}) \).

---

Do you need further details or clarification on any step?

### 5 Related Questions:
1. What is the washer method, and how does it differ from the disk method?
2. How would the volume change if we revolved around the line \( y = 10 \)?
3. How do you compute the volume when revolving around a vertical line instead of a horizontal one?
4. What is the significance of using trigonometric identities in integrals involving \( \sin(x) \) and \( \cos(x) \)?
5. How can the method of cylindrical shells be used for this type of problem?

**Tip**: Always check the limits of integration and the axis of rotation carefully to ensure correct setup of the radius functions in volume problems.

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y = 7. y = sin(x) y = 0 0 <= x <= pi/2 pi(14 - pi/2) ○ pi(28 - pi/4) ○ pi(7 - pi/2) ○ pi(7 - pi/4) pi(14 - pi/4)

Calculate the volume of the solid generated by revolving the region bounded by y = sin(x), y = 0, and 0 <= x <= π/2 about the line y = 7. This problem involves the washer method and trigonometric identities, making it ideal for students in Grades 11-12.

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