Answer:
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!!!
