Use calculus to find the volume of the following solid 
S: 

The base of 
S is the triangular region with vertices (0, 0), (3, 0), and (0, 2). Cross-sections perpendicular to the  
y-axis are semicircles. 

Volume = 

We can use the method of slicing to find the volume of the solid. The given solid \( S \) has a triangular base with vertices at \( (0, 0) \), \( (3, 0) \), and \( (0, 2) \). Cross-sections perpendicular to the \( y \)-axis are semicircles.

### Steps to find the volume:

1. **Equation of the triangle**:  
   The triangular base is bounded by the line connecting the points \( (3, 0) \) and \( (0, 2) \), which has the equation of the line. The slope of the line is:
   \[
   \text{slope} = \frac{2 - 0}{0 - 3} = -\frac{2}{3}
   \]
   Hence, the equation of the line is:
   \[
   x = -\frac{3}{2} y + 3
   \]

2. **Width of the cross-section**:  
   Since cross-sections perpendicular to the \( y \)-axis are semicircles, the diameter of a semicircle at a particular \( y \)-value is the horizontal distance between the points on the triangle. The width of the triangle at height \( y \) is given by \( 2x = 2 \left( -\frac{3}{2} y + 3 \right) = -3y + 6 \).

   The diameter \( d \) of the semicircle is:
   \[
   d = -3y + 6
   \]

3. **Area of the cross-section**:  
   The area \( A(y) \) of a semicircle with diameter \( d \) is:
   \[
   A(y) = \frac{1}{2} \pi \left( \frac{d}{2} \right)^2 = \frac{1}{2} \pi \left( \frac{-3y + 6}{2} \right)^2
   \]
   Simplifying:
   \[
   A(y) = \frac{\pi}{8} \left( -3y + 6 \right)^2
   \]

4. **Volume integral**:  
   To find the volume, we integrate the area function \( A(y) \) from \( y = 0 \) to \( y = 2 \) (the limits of the triangular base along the \( y \)-axis):
   \[
   V = \int_0^2 A(y) \, dy = \int_0^2 \frac{\pi}{8} \left( -3y + 6 \right)^2 \, dy
   \]

5. **Evaluate the integral**:  
   We will expand \( (-3y + 6)^2 \) and then integrate:
   \[
   (-3y + 6)^2 = 9y^2 - 36y + 36
   \]
   So the integral becomes:
   \[
   V = \frac{\pi}{8} \int_0^2 (9y^2 - 36y + 36) \, dy
   \]
   Compute each term:
   \[
   \int_0^2 9y^2 \, dy = 9 \cdot \frac{y^3}{3} \Big|_0^2 = 9 \cdot \frac{8}{3} = 24
   \]
   \[
   \int_0^2 -36y \, dy = -36 \cdot \frac{y^2}{2} \Big|_0^2 = -36 \cdot 2 = -72
   \]
   \[
   \int_0^2 36 \, dy = 36 \cdot y \Big|_0^2 = 36 \cdot 2 = 72
   \]

   Now substitute these results back into the volume expression:
   \[
   V = \frac{\pi}{8} \left( 24 - 72 + 72 \right) = \frac{\pi}{8} \cdot 24 = 3\pi
   \]

### Final Answer:
The volume of the solid is \( V = 3\pi \) cubic units.

---

Would you like further details or have any questions about the steps?  
Here are five related questions you could explore:

1. How would the volume change if the cross-sections were full circles instead of semicircles?
2. What if the cross-sections were squares instead of semicircles?
3. How would you find the volume if the base were a different polygon, like a square or rectangle?
4. What would happen to the integral if the base vertices were translated or rotated?
5. How can we use different methods of integration (e.g., cylindrical shells) to solve this problem?

**Tip:** Understanding the geometric interpretation of cross-sections helps in setting up integrals correctly for volume problems.

Use calculus to find the volume of the following solid S: The base of S is the triangular region with vertices (0, 0), (3, 0), and (0, 2). Cross-sections perpendicular to the y-axis are semicircles.

Learn how to find the volume of a solid with a triangular base and semicircular cross-sections perpendicular to the y-axis using calculus. This problem covers key calculus concepts like integration, cross-sectional area, and the use of definite integrals. Suitable for students in Grades 11-12.

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