MA200 Integral Calculus: Area Between Curves, Volumes of Revolution, and Arc Length
Foundations: Area Between Curves and Riemann Sum Generalization
Riemann Sum Framework: Definite integrals evaluate net signed area. Pure geometric area between curves is always non-negative.
Area Between Curves: For continuous functions with f(x)≥g(x) on [a,b]: A=∫ab(f(x)−g(x))dx
Intersecting Regions: Partition the integral at intersection points to maintain (upper curve)−(lower curve).
Example: Enclosing y=x3 and y=4x over [−1,1] splits at x=0: A=∫−10(x3−4x)dx+∫01(4x−x3)dx=47+47=27
Standard Evaluations:
Enclosed by g(x)=x and f(x)=x over [0,1]: A=∫01(x−x)dx=61
Enclosed by f(x)=4x, g(x)=2x, and x=1: A=∫01(4x−2x)dx=2ln(2)1
Volume by Slicing and Cross-Sectional Area
Slicing Principle: Integrating parallel cross-sectional slices of area A(x) along the x-axis yields total volume: V=∫abA(x)dx
Uniform Cross Section: For a constant 2×2 square cross section on [0,10], V=∫0104dx=40units3.
Square Pyramid: Apex at origin, height h=10, base side b=6. Side length scales as s(x)=53x: A(x)=259x2⟹V=∫010259x2dx=120units3
Right Circular Cone: Apex at origin, height h=10, base radius r=5. Radius scales as r(x)=21x: A(x)=4πx2⟹V=∫0104πx2dx=3250πunits3
Solids of Revolution and the Disk Method
Differentiation Rules and Geometric Foundations
Calculus Rules:
Product Rule: dxd(uv)=u′v+vu′
Quotient Rule: dxd(vu)=v2u′v−v′u
Chain Rule: dxdy=dudydxdu
Exponentials: dxd(ex)=ex, dxd(ax)=ln(a)⋅ax, and dxd(2−x)=−ln(2)⋅2−x
Distance Relations:
Arc Length: Derivation and Rigorous Formulation
Approximation: Segment sum of secant chords: ∑i=1n(Δxi)2+(Δyi)2.
Definite Integral Derivation: Factoring Δxi forms a Riemann sum converging to: L=limn→∞∑i=1n1+(ΔxiΔyi)2Δxi=∫ab1+(f′(x))2dx
Condition: f′(x) must be continuous on [a,b].
Arc Length Computations and Analytical Techniques
Parabola y=21x2 over [0,3]: L=∫031+x2dx≈5.65
Linear function y=2x+1 over [0,2]: L=∫021+22dx=25
Power function y=x3/2 over [0,4] (via u=1+49x): L=∫041+49xdx=278(103/2−1)≈9.07
Logarithmic function f(x)=8x2−ln(x) over [1,2]:
Radicand forms a perfect square: 1+(f′(x))2=(4x+x1)2
Integral evaluation: $$L = \int_1^2 \left(\frac{x}{4} + \frac{1}{x}\right)\,dx = \frac{3}{8} + \ln(2) \approx