Question

If the infinite curve y = e^−3x, x ≥ 0, is rotated about the x-axis, find the area of the resulting surface.

Answer 1

Answer:

S = 2π/9 [³/₂√10 − ½ ln(-3 + √10)]

S ≈ 3.946

Step-by-step explanation:

For a curve rotated about the x-axis, the surface area is:

S = ∫ₐᵇ 2πy ds,

where ds = √(1 + (dy/dx)²) dx.

y = e⁻³ˣ

dy/dx = -3e⁻³ˣ

ds = √(1 + (-3e⁻³ˣ)²) dx

S = ∫₀°° 2πe⁻³ˣ √(1 + (-3e⁻³ˣ)²) dx

If u = -3e⁻³ˣ, then du = 9e⁻³ˣ dx, or du/9 = e⁻³ˣ dx.

When x = 0, u = -3.  When x = ∞, u = 0.

S = ∫₋₃⁰ 2π √(1 + u²) (du/9)

S = 2π/9 ∫₋₃⁰ √(1 + u²) du

S = 2π/9 [ ½ u √(1 + u²) + ½ ln|u + √(1 + u²)| ] |₋₃⁰

S = 2π/9 {[0] − [ -³/₂√10 + ½ ln(-3 + √10) ]}

S = 2π/9 [³/₂√10 − ½ ln(-3 + √10)]

S ≈ 3.946