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)) ]

step-by-step explanation:

consider an orange peel: if you want to try and lay it flat, you have to stretch it, we know exactly how things are being stretched or squashed at any given point. we have many different map projections because each has different patterns of some projections can even preserve certain features of the earth without