University:
Johns Hopkins University
Course:
AS.110.601 | Algebra I
Academic year:

2024
Views:

85

Pages:

6
Author:

Cristian Santos

Question 1 Evaluate the line integral, where C is the given curve. ∫C xy ds C: x = t², y = 2t, 0 ≤ t ≤ 4 Answer Step 1 Consider x = t² dx/dt = 2t y = 2t dy/dt = 2 Consider: ds = √( (dx/dt)² + (dy/dt)² ) dt = √( (2t)² + (2)² ) dt = √(4t² + 4) dt = √(4(t² + 1)) dt ds = 2√(t² + 1) dt Hence the line integral is, ∫C xy ds = ∫₀⁴ (t²)(2t)(2√(t² + 1)) dt = 4 ∫₀⁴ t³√(t² + 1) dt ... (1) Step 2 Let t² + 1 = u² 2t dt = 2u du t dt = u du For t = 0 then t² + 1 = u² u² = 1 u = 1 For t = 4 then t² + 1 = u² u² = 4² + 1 u² = 17 u = √17 Substitute these values in (1): ∫C xy ds = 4 ∫₁√¹⁷ t³√(t² + 1) dt = 4 ∫₁√¹⁷ √(u²) × u du = 4 ∫₁√¹⁷ (u² - 1)u du = 4 ∫₁√¹⁷ (u⁴ - u²) du = 4 [u⁵/5 - u³/3]₁√¹⁷ = 4 [(√17)⁵/5 - (√17)³/3 - (⅕ - ⅓)] = 4 [ (782√17 / 15) - (2/15) ] ∫C xy ds = 860.33 Question 2 Use a double integral to find the volume of the solid shown in the figure. Answer 1 Solution: V = ∫₀³ ∫₀⁶ (6 - y) dy dx Answer 2 Solution: V = ∫₀³ ∫₀⁶ (6 - 4) dy dx V = ∫₀³ ∫₀⁴ 6 dy dx V = ∫₀³ (6(4)) dx V = ∫₀³ 16 dx V = 16(3) = 48 Question 3 Use geometry or symmetry, or both, to evaluate the double integral. ∬D √(R² - x² - y²) dA D is the disk with center the origin and radius R. Answer ∬D √(R² - x² - y²) dA We can write z = √(R² - x² - y²) which can become x² + y² = R² - z², z ≥ 0 Thus, z is the upper half, a hemisphere. We find the volume using the sphere formula, taking half: V = ½ (4/3 πR³) Results: 2/3 πR³ Question 4 Evaluate the integral by making the given substitution. ∫ x³ (3 + x⁴)⁷ dx, u = 3 + x⁴ Answer Step 1 x³(3 + x⁴)⁷ dx, u = 3 + x⁴ Substitute (3 + x⁴) = u 4x³ dx = ∂u x³ dx = ∂u/4 ∫ x³ (u⁷) ∂u/4 = (u⁸ / 4 × 8) + c Step 2 = 1/32 (3 + x⁴)⁸ + c Question 5 How to prove ∫₀∞ e⁻ˣ² dx = √π/2 Answer 1 This is an old favorite. Define I = ∫₀∞ e⁻ˣ² dx Then I² = (∫₀∞ e⁻ˣ² dx)(∫₀∞ e⁻ʸ² dy) I² = ∫₀∞ ∫₀∞ e⁻(x² + y²) dx dy Now switch to polar coordinates: I² = ∫₀²π ∫₀∞ e⁻ʳ² r dr dθ The θ integral gives 2π: I² = 2π ∫₀∞ e⁻ʳ² r dr Using substitution, u = r²: I² = π I = √π Answer 2 We can observe: (∫₀∞ e⁻ˣ² dx)(∫₀∞ e⁻ʸ² dy) = V Volume of the body is bounded as a solid of revolution: V = ∫₀¹ π(-ln z) dz = π

Integrals #4 - Questions and Answers

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