Answer: the area of the resulting surface is π/16 [ 4√(17) + In( 4 + √( 17)) ]
Explanation:
Given that;
y = e^(-4x), x ≥ 0
then, dy/dx = -4e^(-4x)
now, √( 1 + (dy/dx)² ) = √( 1 + ( -4e^(-4x))²) = √( 1 + 16e^(-8x)
so the surface area of the surface obtaining by rotating the curve y = e^(-4x) on ( 0, ∞ ) about x-axis is,
S = ₀∫^∞ 2πy √( 1 + (dy/dx)² ) dx
S = ₀∫^∞ 2πe^(-4x) . √( 1 + 16e^(-8x) dx
S = 2π. ₀∫^∞ e^(-4x) . √( 1 + 16e^(-8x) dx
so LET 4e^-4x = u ⇒ -16e^(-4x) dx = du
⇒e^-4x dx = du/ -16
so whenx = 0, then u =4 and when x → ∞ then e^(-4x) → 0 so u →0
therefore
S = 2π ₄∫⁰ √(1 + u²) . du/-16
= π/8 ₀∫⁴ √(1 + u²) . du
= π/8 [ u/2√(1 + u²) + 1/2 In ( u + √( 1 + u²) ]₀⁴
WE MAKE USE OF
∫√( a² + u²) du = u/2√(a² + u²) + a²/2 In ( u + √( a² + u²)) + C
so
= π/8 [ 4/2√(1 + 16) + 1/2 In ( 4 + √( 1 + u16 ]
= π/8 [ 2√(17) + 1/2 In ( 4 + √( 17)) ]
= π/16 [ 4√(17) + In( 4 + √( 17)) ]
therefore the area of the resulting surface is π/16 [ 4√(17) + In( 4 + √( 17)) ]