Finding the Volume of a Solid Using Cross Sections

Key Concepts

• Two main formulas for finding the volume of a solid through integration:
  1. Integral from a to b of the area function A(x) dx (cross-sections perpendicular to the x-axis).
  2. Integral from c to d of the area function A(y) dy (cross-sections perpendicular to the y-axis).

Problem 1: Volume with Square Cross Sections

• Function & Boundaries: y = sqrt(x), bounded by the x-axis and the line x = 4
• Cross Sections: Squares, perpendicular to the x-axis
• Area of Cross Section: Side^2 (s^2)
  • s is the base of the cross-section
  • y = s
  • A(x) = y^2 = (sqrt(x))^2 = x
• Volume Calculation:
  • Integral from 0 to 4 of x dx
  • Antiderivative: x^2/2 from 0 to 4
  • Calculation: 16/2 - 0 = 8

Problem 2: Volume with Semi-Circular Cross Sections

• Function & Boundaries: y = 4 - x/2, bounded by the x-axis, y-axis
• Cross Sections: Semi-circles, perpendicular to the x-axis
• Graph:
  • Line: Slope -1/2, y-intercept 4
  • Equation rewritten as: y = -1/2 x + 4
  • X-intercept: x = 8
• Area of Cross Section:
  • Diameter s = y
  • Radius r = s/2
  • Area of semi-circle: 1/2 π r^2 = 1/2 π (s/2)^2 = 1/2 π (1/4 s^2) = 1/8 π s^2
  • s = y = 4 - x/2
  • A(x) = 1/8 π (4 - x/2)^2
• Volume Calculation:
  • Integral from 0 to 8: 1/8 π (4 - x/2)^2 dx
  • Expand A(x): (4 - x/2)^2 = 16 - 4x + 1/4 x^2
  • Integral: 1/8 π ∫ (16 - 4x + 1/4 x^2) dx
  • Separate constants and integrate: 1/8 π (16x - 4x^2/2 + x^3/12) from 0 to 8
  • Evaluate: 1/8 π ([8^3/12 + 8^2/2 * 2 + 16 * 8] - 0)
  • Simplify: 1/8 π ([64/12 + 32 + 128])
  • Solution: 16π/3 ≈ 16.755

Summary

• Utilize geometric properties of cross-sections to find area functions in terms of x or y.
• Apply definite integrals to compute the volume.
• Simplify by recognizing relationships between cross sections, their areas, and integration limits.