Find the exact area of the surface obtained by rotating the curve about the x-axis.

Question

Find the exact area of the surface obtained by rotating the curve about the x-axis.

y=sin((pi*x)/5), 0<=x<=5

Answer 1

The surface area of a curve rotated about an axis is governed by the following formula:

S = ∫ 2π y ds , y = f(x) = sin( πx/5) , ds = √ 1+(dy/dx)² dx , dy/dx = π/5 cos( πx/5)

S = ∫ 2π sin(πx/5) √ 1 + ( π/5 cos( πx/5)² dx , now solve the integration

let π/5 cos( πx/5) = u , du = -π/5( sin(πx/5) dx, and -5/π du = ( sin(πx/5) dx , substitute for u

S = -5/π ∫ 2 π √1+u² du = -10 ∫ √1+u² du

S= -10 ∫ √ 1+u² du to solve this you need to use trigonometric substitution of the form √ a² + u²

Hera a = 1, let u = a tan θ = tan θ , du = a sec²θ dθ = sec²θ dθ

S = -10 ∫ √ 1 + tan²θ sec²θ dθ

knowing that 1+ tan²θ = sec²θ from trigonometric identities

S = -10 ∫ √sec²θ sec²θ dθ = -10 ∫ secθ sec²θ dθ = -10 ∫ sec³θ dθ solving this integration

S = -10 { 1/2( secθ tanθ ) + ln [ secθ + tanθ] } with bounds from 0 to 5

we know that u is tanθ therefore from a right angle triangle secθ = √ 1 + u²

Then S = -10 { 1/2( u √ 1 + u² ) + Ln [ √ 1 + u² + u ] } now substitute for u = π/5 cos( πx/5)

S = -10 { 1/2( π/5 cos( πx/5) √ 1 + ( π/5 cos( πx/5)² + Ln [ √ 1 + ( π/5 cos( πx/5)² + π/5 cos( πx/5) }

now substitute for the bounds 0 and 5

for x = 0, π/5 cos( πx/5) = π/5 cos(0) = π/5

for x = 5 π/5 cos( πx/5) = π/5 cos( π) = -π /5

S = -10 { 1/2( π/5)² √ 1 + ( π²/25)² + Ln [ √ 1 + ( π²/25)² + π/5 ) - 1/2( π/5)(-π/5) √ 1 + [( π/5)(-π/5)]² - Ln [ √ 1 + [( π/5)(-π/5)]² + π/5 ] } the first and the third term add up and the Ln terms cancel out

S = -10 { 2 (1/2) ( π/5)² √ 1 + ( π²/25)² } , s = -10 { ( π/5)² √ 1 + ( π²/25)² } solving

S = -10 { 0.42} = - 4.2 unit ² , the negative term can be dropped , and S = 4.2 unit ²