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Volumes With Cross Sections: AP Calculus AB Rundown

This topic focuses on finding the volume of a 3D solid based on knowing the area of its cross-sections perpendicular to a given axis.

1. General Principle

The fundamental idea is to slice the solid into infinitesimally thin cross-sections, calculate the area of each cross-section, and then integrate the area function over the interval of the solid.

$$ V = \int_{a}^{b} A(x) , dx \quad \text{or} \quad V = \int_{c}^{d} A(y) , dy $$
Where:

• $ V $ is the volume of the solid.
• $ A(x) $ or $ A(y) $ is the area of the cross-section at position $ x $ or $ y $ , respectively.
• $ [a, b] $ or $ [c, d] $ are the intervals on the x-axis or y-axis, respectively, over which the solid extends.

2. Key Steps to Solve Volume with Cross Sections Problems

1. Sketch the Base: Carefully sketch the region that forms the base of the solid. This will help you determine the limits of integration and the relationship between $ x $ and $ y $ .
2. Determine the Axis of Integration: Decide whether to integrate with respect to $ x $ or $ y $ based on the orientation of the cross-sections. If the cross-sections are perpendicular to the x-axis, integrate with respect to $ x $ . If they are perpendicular to the y-axis, integrate with respect to $ y $ .
3. Find the Area Function A(x) or A(y): This is the most crucial step. You need to express the area of the cross-section as a function of $ x $ or $ y $ . The specific formula for $ A(x) $ or $ A(y) $ depends on the shape of the cross-section. Area Formulas for Common Shapes
4. Determine the Limits of Integration: Find the values of $ x $ (or $ y $ ) that define the beginning and end of the solid along the chosen axis.
5. Integrate: Evaluate the definite integral of the area function over the limits of integration to find the volume.

3. Common Cross-Section Shapes and Their Area Formulas

Here are some common shapes you might encounter, along with their area formulas:

• Squares: If the side length of the square is $ s $ , then $ A = s^2 $ .
• Rectangles: If the length is $ l $ and the width is $ w $ , then $ A = lw $ . Usually one dimension is constant and the other is defined by the region.
• Semicircles: If the diameter is $ d $ , then the radius is $ r = d/2 $ , and $ A = \frac{1}{2}\pi r^2 = \frac{1}{2}\pi (\frac{d}{2})^2 = \frac{\pi}{8}d^2 $ .
• Equilateral Triangles: If the side length is $ s $ , then $ A = \frac{\sqrt{3}}{4}s^2 $ .
• Isosceles Right Triangles: If the leg length is $ l $ , then $ A = \frac{1}{2}l^2 $ .
• Right Triangles: if the base is $ b $ and height is $ h $ , then $ A = \frac{1}{2}bh $

In all these cases, the side length, diameter, or base/height will be expressed in terms of $ x $ or $ y $ based on the region defining the base of the solid.

4. Examples

Example 1:

The base of a solid is the region bounded by $ y = x^2 $ and $ y = 4 $ . Cross-sections perpendicular to the x-axis are squares. Find the volume.

1. Sketch: Sketch the parabola $ y = x^2 $ and the line $ y = 4 $ .

2. Axis of Integration: Since the cross-sections are perpendicular to the x-axis, we integrate with respect to $ x $ .

3. Area Function: The side length of the square is the distance between the two curves: $ s = 4 - x^2 $ . Therefore, the area of the square is $ A(x) = (4 - x^2)^2 $ .

4. Limits of Integration: Find the points of intersection: $ x^2 = 4 $ , so $ x = \pm 2 $ . The limits of integration are $ -2 $ and $ 2 $ .

5. Integrate:

   $$ V = \int_{-2}^{2} (4 - x^2)^2 , dx = \int_{-2}^{2} (16 - 8x^2 + x^4) , dx $$
   $$ V = \left[16x - \frac{8}{3}x^3 + \frac{1}{5}x^5\right]_{-2}^{2} = 2\left(32 - \frac{64}{3} + \frac{32}{5}\right) = \frac{512}{15} $$