Calculus 1 — 28.3: The Washer Method скачать в хорошем качестве

Calculus 1 — 28.3: The Washer Method

When a solid of revolution has a hollow center — like a vase, pipe, or toilet paper roll — the disc method falls short. The washer method extends the disc approach by subtracting the inner radius from the outer radius, letting you compute volumes of hollow solids in a single integral. This video walks through the derivation of the washer formula, highlights the most common algebraic mistakes, and solves two complete examples step by step. Key concepts covered: • Why the disc method fails for solids with hollow centers • Deriving the washer area formula: A(x) = π([f(x)]² − [g(x)]²) • The volume integral: V = π∫ from a to b of ([f(x)]² − [g(x)]²) dx • Identifying outer vs. inner functions using a test point • Finding limits of integration from intersection points • Critical mistake: why you must square each radius individually, not subtract first then square • Why π∫(f−g)² dx gives the wrong answer (unwanted cross term −2fg) • Why π∫(f−g) dx gives area between curves, not volume • Advantage of one combined integral over two separate disc integrals (like terms cancel) • Worked Example 1: f(x) = x² + ½ and g(x) = x on [0, 2], yielding V = 69π/10 • Worked Example 2: y = x² and y = x³ revolved around the x-axis, yielding V = 2π/35 • Five-step washer method workflow: bounds, outer/inner check, integral setup, simplification, evaluation • Challenge problem: y = √x and y = x² on [0, 1], with answer V = 3π/10 ━━━━━━━━━━━━━━━━━━━━━━━━ SOURCE MATERIALS The source materials for this video are from    • Calculus 1 Lecture 5.2:  Volume of Solids ...