1. Find the area of the surface obtained by revolving y = sin(x), 0 ≤ x ≤ π, about the x-axis.
Answer
1. The area is 2π(√2 + (1/2)ln(1 + √2) – (1/2)ln(-1 + √2))
Solution
1. Because f(x) = sin(x), f'(x) = cos(x) and the area is ∫0π2π sin(x) √(1 + cos2(x)) dx = -2π∫1- 1√(1 + u2)du substituting u = cos(x) = -2π((u/2)√(1 + u2) + (1/2)ln(u + √(1 + u2)))|1-1 by 16 from the Integral Table = 2π(√2 + (1/2)ln(1 + √2) – (1/2)ln(-1 + √2))
2. Find the area of the surface obtained by revolving y = √x, 0 ≤ x ≤ 1, about the x-axis.
Answer
1. The area is (5√5 – 1)/12.
Solution
1. Because f(x) = √x, f ‘(x) = 1/(2√x) and the area is ∫01 2π √x √(1 + (1/(2√x))2) dx = 2π∫01√(x + 1/4) dx = 2π (2/3)(x + 1/4)3/2|01 = (5√5 – 1)/12
3. Find the area of the surface obtained by revolving y = ex, 0 ≤ x ≤ 1, about the x-axis.
Answer
1. The area is -2π (((1/2e)√(1 + 1/e2) + (1/2)ln(1/e + √(1 + 1/e2)) – (√2/2 + (1/2)ln(1 + √2)))
Answer
1. Because f(x) = e-x, f'(x) = -e-x and the area is ∫01 2π e-x √(1 + (-e-x)2) dx = -2π∫11/e√(1 + u2) du substituting u = e-x = -2π((u/2)√(1 + u2) + (1/2)ln(u + √(1 + u2)))|11/e by 16 from the Integral Table = -2π (((1/2e)√(1 + 1/e2) + (1/2)ln(1/e + √(1 + 1/e2)) – (√2/2 + (1/2) ln(1 + √2)))
4. Find the area of the surface obtained by revolving y = √(x+1), 0 ≤ x ≤ 1, about the x-axis.
Answer
1. The area is π(27 – 5√5)/6.
Solution
1. Because f(x) = √(x+1), f ‘(x) = 1/(2√(x+1)) and the area is ∫01 2π √(x+1) √(1 + (1/(2√(x+1)2) dx = 2π∫01√(x + 1 + 1/4) dx = 2π (2/3)(x + 5/4)3/2|01 = π(27 – 5√5)/6
5. Find the area of the surface obtained by revolving y = √(2x – x2), 0 ≤ x ≤ 1, about the x-axis.
Answer
1. The area is 2π.
Solution
1. Because f(x) = √(2x – x2), f ‘(x) = (1 – x)/√(2x – x2) and the area is ∫01 2π √(2x – x2) √(1 + ((1 – x)/√(2x – x2)2) dx = 2π∫01 √(2x – x2 + (1 – x)2) dx = 2π∫01 √1 dx = 2π