1. Rotating x= g(y) around the y-axis gives a circle of radius g(y), so the circumference 2πf(x) in the formula for revolving about the x-axis is replaced by the circumference 2π g(y). Similarly, the arclength factor √(1 + (f ‘(x)²) is replaced by the arclength factor √(1 + (g ‘(y)²). The result follows by combining these factors.

1. The area is (π/27)(10√10 – 1)

1. Because g(y)= y³, g ‘(y) = 3y² and the area is: ∫₀¹2π y³ √(1 + (3y²)²) dy = 2π∫₀¹y³√(1 + 9y⁴) dy = 2π(1/36)(2/3)u^3/2|₁¹⁰ by substituting u= 1 + 9y⁴ = (π/27)(10√10 – 1)

1. Because g(y) = y⁴, g ‘(y) = 4y³ and the area is ∫₀²2π y⁴ √(1 + (4y³)²) dy = 2π∫₀² y⁴ √(1 + 16y⁶) dy No obvious – or non-obvious, as far as we can tell – trick works to evaluate this integral. But notice that the terms inside the square root are always ≥ 1, so area ≥ 2π∫₀² y⁴ dy = 64π/5.

1. Because g(y) = 1/y, g'(y)= -1/y² and the area is: ∫₁²2π 1/y √(1 + (-1/y²)²) dy = 2π∫₁² 1/y √(1 + 1/y⁴) dy No obvious – or non-obvious, as far as we can tell – trick works to evaluate this integral. But notice that the terms inside the square root are always ≥ 1, so area ≥ 2π∫₁²1/y dy = 2π ln(y)|₁² = 2π ln(2).

1. The value of a= ((55^2/3 – 1)/9)^1/4 ≈ 1.106.

1. Because f(x)= x³, f'(x)= 3x² and the area is: ∫₀^a2π x³ √(1 + 9x⁴) dx = 2π((1 + 9a⁴)^3/2 – 1)/54 Setting area = 2π and solving for a gives a = ((55^2/3 – 1)/9)^1/4.