Answer:
To compute the integral ∫γω along the path γ, we need to parameterize the path γ and then evaluate the integral using the parameterization.
The path γ traces the graph of the function g(x) = sin(x) on the interval [0, π]. We can parameterize this path as follows:
x = t (where t ∈ [0, π])
y = sin(t)
Now, let's compute the differential forms dx and dy:
dx = dt
dy = cos(t) dt
Substituting these expressions into ω = (cos(x) + y) dx + (4x + 4y³) dy, we get:
ω = (cos(t) + sin(t)) dt + (4t + 4sin(t)³) cos(t) dt
= (cos(t) + sin(t) + (4t + 4sin(t)³) cos(t)) dt
Now, we can evaluate the integral along the path γ:
∫γ ω = ∫₀ᴨ (cos(t) + sin(t) + (4t + 4sin(t)³) cos(t)) dt
To evaluate this integral, we can split it into three separate integrals:
I₁ = ∫₀ᴨ cos(t) dt
I₂ = ∫₀ᴨ sin(t) dt
I₃ = ∫₀ᴨ (4t + 4sin(t)³) cos(t) dt
Evaluating each integral separately:
I₁ = [sin(t)]₀ᴨ = sin(ᴨ) - sin(0) = 0 - 0 = 0
I₂ = [-cos(t)]₀ᴨ = -cos(ᴨ) + cos(0) = -(-1) + 1 = 2
To evaluate I₃, we can use integration by parts:
Let u = 4t + 4sin(t)³ and dv = cos(t) dt
Then du = 4 + 12sin(t)² cos(t) dt and v = sin(t)
Using the formula for integration by parts, we have:
I₃ = uv - ∫v du
= (4t + 4sin(t)³) sin(t) - ∫sin(t) (4 + 12sin(t)² cos(t)) dt
= (4t + 4sin(t)³) sin(t) - ∫(4sin(t) + 12sin(t)³ cos(t)) dt
= (4t + 4sin(t)³) sin(t) - (4∫sin(t) dt + 12∫sin(t)³ cos(t) dt)
= (4t + 4sin(t)³) sin(t) - (-4cos(t) - 12I₃)
Simplifying the equation:
I₃ = (4t + 4sin(t)³) sin(t) + 4cos(t) + 12I₃
11I₃ = (4t + 4sin(t)³) sin(t) + 4cos(t)
I₃ = ((4t + 4sin(t)³) sin(t) + 4cos(t))/11
Now, substituting the values of I₁, I₂, and I₃ back into the original integral:
∫γ ω = I₁ + I₂ + I₃
= 0 + 2 + ((4t + 4sin(t)³) sin(t) + 4cos(t))/11
Therefore, the value of the integral ∫γω along the path γ is ((4t + 4sin(t)³) sin(t) + 4cos(t))/11 + 2.
Explanation: