If the infinite curve y = e−4x, x ≥ 0, is rotated about the x-axis, find the area of the resulting surface.

the area of the resulting surface is π/16 [ 4√(17)  + In( 4 + √( 17)) ]

Step-by-step 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)) ]

whats the question?

step-by-step explanation: