Application of Integrals

Integration transforms from abstract mathematical operation to practical tool in this chapter. While geometry taught you to calculate areas of simple shapes (rectangles, triangles, circles), those formulas fail for irregular regions bounded by curves. Integration solves this: by summing infinitesimal rectangular strips under a curve, we calculate areas that simple geometry cannot touch. Beyond area, integrals compute volumes of complex 3D solids, lengths of curves, and centers of mass. This chapter applies the integration techniques from chapter-07-integrals to real geometric and physical problems, bridging to chapter-11-three-dimensional-geometry.

Area Under a Curve

The fundamental problem: Find the area bounded by y = f(x), the x-axis, and vertical lines x = a and x = b.

Solution: If f(x) ≥ 0 on [a, b], then:

Area = ∫ₐᵇ f(x) dx

Intuition: Divide the region into n thin vertical strips. Each strip has width Δx and height approximately f(xᵢ). The total area is approximately Σf(xᵢ)·Δx. As n → ∞ and Δx → 0, this becomes the integral.

Example: Area under y = x² from x = 0 to x = 2:

Area = ∫₀² x² dx = [x³/3]₀² = 8/3 - 0 = 8/3

Area Between Two Curves

For curves y = f(x) and y = g(x) where f(x) ≥ g(x) on [a, b]:

Area = ∫ₐᵇ [f(x) - g(x)] dx

Strategy:

1. Sketch both curves to identify the upper and lower curves
2. Find intersection points (where f(x) = g(x)) to determine integration bounds
3. Integrate the difference over the appropriate regions

Example: Area between y = x² and y = x from x = 0 to x = 1:

• Curve y = x is above y = x² on [0, 1]
• Area = ∫₀¹ (x - x²) dx = [x²/2 - x³/3]₀¹ = 1/2 - 1/3 = 1/6

Volumes of Solids of Revolution

Rotating a 2D region around an axis creates a 3D solid. Two methods calculate volume:

Disk Method (rotation around x-axis or y-axis)

When rotating the region under y = f(x) around the x-axis from x = a to x = b:

Volume = π ∫ₐᵇ [f(x)]² dx

Intuition: At each x, the cross-section is a disk with radius f(x) and area π[f(x)]². Integrate these disk areas along the x-axis.

Example: Rotate y = √x from x = 0 to x = 4 around the x-axis:

V = π ∫₀⁴ (√x)² dx = π ∫₀⁴ x dx = π[x²/2]₀⁴ = 8π

Washer Method (for hollow solids)

When the region is between y = f(x) and y = g(x) with f(x) > g(x) ≥ 0, rotating around x-axis:

Volume = π ∫ₐᵇ {[f(x)]² - [g(x)]²} dx

Intuition: Each cross-section is a washer (disk with hole) with outer radius f(x) and inner radius g(x). Area of washer is π[f(x)]² - π[g(x)]².

Shell Method (alternative approach)

For rotation around the y-axis, cylindrical shells often simplify calculation:

Volume = 2π ∫ₐᵇ x·f(x) dx

Intuition: Imagine wrapping the region in concentric cylindrical shells. Each shell at radius x has height f(x) and thickness dx. The shell's lateral surface area is 2πx (circumference) times the height f(x).

Volume of Solids with Known Cross-Sections

If cross-sectional area perpendicular to an axis is known as a function A(x):

Volume = ∫ₐᵇ A(x) dx

Example: A pyramid with square base (side 2) and height 3. At height x from the apex, the square has side length (2x/3). Cross-sectional area A(x) = (2x/3)² = 4x²/9.

Volume = ∫₀³ (4x²/9) dx = (4/9)[x³/3]₀³ = (4/9) · 9 = 4

Arc Length

The length of a curve y = f(x) from x = a to x = b (where f'(x) is continuous):

Arc Length = ∫ₐᵇ √[1 + (f'(x))²] dx

Intuition: Small segment ds ≈ √[1 + (dy/dx)²] dx (using Pythagorean theorem on the tangent line). Integrate these segments.

Example: Arc length of y = x^(3/2) from x = 0 to x = 4:

• f'(x) = (3/2)x^(1/2)
• Arc Length = ∫₀⁴ √[1 + (9/4)x] dx (requires substitution)

Connections to Other Topics

• chapter-07-integrals: Integration techniques form the foundation
• chapter-06-application-of-derivatives: Rate of change relates to arc length
• chapter-11-three-dimensional-geometry: 3D geometry and coordinates
• chapter-10-vector-algebra: Vector approach to area and volume calculations

Socratic Questions

1. Explain how the disk method for volumes of revolution extends the concept of area under a curve. Why does the radius of each disk become [f(x)], and why is it squared in the formula?

1. When rotating a region around the x-axis using the washer method, the volume formula contains [f(x)]² - [g(x)]². Why is subtraction necessary here, whereas finding the area between curves requires f(x) - g(x)?

1. The shell method and disk method can both compute the volume of a solid of revolution. When would you prefer shells over disks? How does the geometry of the region guide your choice?

1. Consider rotating the region bounded by y = x, the x-axis, and x = 2 around the y-axis. Explain why the shell method is more natural here than the disk method. What would be the challenge with using disks?

1. For arc length, why does the integrand contain √[1 + (f'(x))²]? How does this formula relate to the Pythagorean theorem applied to an infinitesimal segment of the curve?