Write an integral for the area of the surface generated by revolving the curve y equals cosine (2 x )about the x-axis on negative StartFraction pi Over 5 EndFraction less than or equals x less than or equals StartFraction pi Over 5 EndFraction .

The integral is

∫ˣ²ₓ₁ 2π cos 2x √[1 + 4 sin² 2x] dx

x₁ = (-π/5)

x₂ = (π/5)

And the area of the surface generated by revolving = 9.71 square units

Step-by-step explanation:

When a function y = f(x) is revolved about the x-axis, the formula for the area of the surface generated is given by

A = 2π ∫ˣ²ₓ₁ f(x) √[1 + (f'(x))²] dx

A = 2π ∫ˣ²ₓ₁ y √[1 + y'²] dx

For this question,

y = cos 2x

x₁ = (-π/5)

x₂ = (π/5)

y' = -2 sin 2x

1 + y'² = 1 + (-2 sin 2x)² = (1 + 4 sin² 2x)

So, the Area of the surface of revolution is

A = 2π ∫ˣ²ₓ₁ y √[1 + y'²] dx

= ∫ˣ²ₓ₁ 2πy √[1 + y'²] dx

Substituting these variables

A = ∫ˣ²ₓ₁ 2π cos 2x √[1 + 4 sin² 2x] dx

Let 2 sin 2x = t

4 cos 2x dx = dt

2 Cos 2x dx = (dt/2)

dx = (1/2cos 2x)(dt/2)

Since t = 2 sin 2x

when x = (-π/5), t = 2 sin (-2π/5) = -1.90

when x = (π/5), t = 2 sin (2π/5) = 1.90

A

= ∫¹•⁹⁰₋₁.₉₀ π (2 Cos 2x) √(1 + t²) (1/2cos 2x)(dt/2)

= ∫¹•⁹⁰₋₁.₉₀ (π/2) √(1 + t²) (dt)

= (π/2) ∫¹•⁹⁰₋₁.₉₀ √(1 + t²) (dt)

But note that

∫ √(a² + x²) dx

= (x/2) √(a² + x²) + (a²/2) In |x + √(a² + x²)| + c

where c is the constant of integration

So,

∫ √(1 + t²) dt

= (t/2) √(1 + t²) + (1/2) In |t + √(1 + t²)| + c

∫¹•⁹⁰₋₁.₉₀ √(1 + t²) (dt)

= [(t/2) √(1 + t²) + (1/2) In |t + √(1 + t²)|]¹•⁹⁰₋₁.₉₀

= [(1.90/2) √(1 + 1.90²)+ 0.5In |1.90+√(1 + 1.90²)|] - [(-1.9/2) √(1 + -1.9²) + (1/2) In |-1.9 + √(1 + -1.9²)|]

= [(0.95×2.147) + 0.5 In |1.90 + 2.147|] - [(-0.95×2.147) + 0.5 In |-1.90 + 2.147|]

= [2.04 + 0.5 In 4.047] - [-2.04 + 0.5 In 0.247]

= [2.04 + 0.70] - [-2.04 - 1.4]

= 2.74 - [-3.44]

= 2.74 + 3.44

= 6.18

Area = (π/2) ∫¹•⁹⁰₋₁.₉₀ √(1 + t²) (dt)

= (π/2) × 6.18

= 9.71 square units.

Hope this Helps!!!

many reason,

step-by-step explanation: