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The problem provided involves calculating the volume of an object where semicircular cross-sections are perpendicular to the base, which is outlined in the \( xy \)-plane. To compute the volume of this object, we need to follow these steps:

### Step-by-step solution:

1. **Identify the base of the object:**
   The base is the blue shape in the \( xy \)-plane, as shown on the left side of the image. It spans from \( x = -1 \) to \( x = 1 \) along the \( x \)-axis, and the height of the shape reaches from \( y = 0 \) to \( y = 6 \). 

2. **Find the equation of the shape:**
   - From \( x = -1 \) to \( x = 1 \), the shape resembles a trapezoid with vertical boundaries. 
   - For the height \( h(y) \) at each \( y \)-value, the width at the bottom is 0 at \( y = 0 \), and it reaches a maximum of 2 units wide (from \( x = -1 \) to \( x = 1 \)) at \( y = 6 \).
   
   The width of the base at a given height \( y \) is proportional to \( y \). The function describing the width is:
   \[
   w(y) = \frac{y}{3} \quad \text{(since the width increases from 0 to 2 as \( y \) goes from 0 to 6)}.
   \]
   
3. **Volume formula using semicircular cross-sections:**
   For each semicircular cross-section, the diameter is the width \( w(y) \). The radius \( r(y) \) is half of this width:
   \[
   r(y) = \frac{w(y)}{2} = \frac{y}{6}.
   \]
   The area \( A(y) \) of a semicircle with radius \( r(y) \) is given by:
   \[
   A(y) = \frac{1}{2} \pi r(y)^2 = \frac{1}{2} \pi \left( \frac{y}{6} \right)^2 = \frac{\pi}{72} y^2.
   \]
   
4. **Integrate to find the volume:**
   The volume of the solid is found by integrating the area of the cross-sections along the \( y \)-axis, from \( y = 0 \) to \( y = 6 \):
   \[
   V = \int_0^6 A(y) \, dy = \int_0^6 \frac{\pi}{72} y^2 \, dy.
   \]
   
5. **Solve the integral:**
   Now we compute the integral:
   \[
   V = \frac{\pi}{72} \int_0^6 y^2 \, dy = \frac{\pi}{72} \left[ \frac{y^3}{3} \right]_0^6 = \frac{\pi}{72} \cdot \frac{6^3}{3} = \frac{\pi}{72} \cdot \frac{216}{3} = \frac{\pi}{72} \cdot 72 = \pi.
   \]

Thus, the exact value of the volume is:

\[
V = \pi.
\]

Would you like further details or have any questions?

### Related questions:
1. How do we derive the formula for the area of a semicircle?
2. What would the volume be if the cross-sections were full circles instead of semicircles?
3. Can this method be applied if the cross-sections are equilateral triangles instead of semicircles?
4. How does the width function \( w(y) \) change if the shape is a different polygon?
5. What would happen to the volume if the shape were rotated around the \( y \)-axis?

### Tip:
When dealing with solids of known cross-sections, always start by finding an expression for the area of the cross-section in terms of the axis you're integrating along.

Compute the exact volume of an object with a given base in the xy-plane and semicircular cross-sections perpendicular to the base.

Learn how to compute the exact volume of a solid with a given base in the xy-plane and semicircular cross-sections perpendicular to the base. This step-by-step solution involves key calculus and geometry concepts suitable for Grades 11-12.

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