Evaluate the line integral, where C is the given curve.
C z dx + x dy + y dz, C: x = t^4, y = t^5, z = t^4, 0 ≤ t ≤ 1
To evaluate the line integral, we need to integrate the given expression along the given curve C.
Let's start by parameterizing the curve C using the given equations:
x = t^4
y = t^5
z = t^4
Next, we need to find the differentials dx, dy, and dz in terms of dt. Taking the derivatives of the parametric equations, we have:
dx = 4t^3 dt
dy = 5t^4 dt
dz = 4t^3 dt
Substituting these expressions into the line integral, we get:
∫(C) z dx + x dy + y dz = ∫(0 to 1) (t^4)(4t^3 dt) + (t^4)(5t^4 dt) + (t^5)(4t^3 dt)
Simplifying the integrand, we have:
∫(C) z dx + x dy + y dz = ∫(0 to 1) 4t^7 dt + 5t^8 dt + 4t^8 dt
Now, we can integrate term by term:
∫(C) z dx + x dy + y dz = ∫(0 to 1) (4t^7 + 5t^8 + 4t^8) dt
Combining like terms, we have:
∫(C) z dx + x dy + y dz = ∫(0 to 1) (4t^7 + 9t^8) dt
Integrating each term, we get:
∫(C) z dx + x dy + y dz = [4/8 t^8 + 9/9 t^9] evaluated from 0 to 1
Evaluating at the upper and lower limits:
∫(C) z dx + x dy + y dz = (4/8 (1)^8 + 9/9 (1)^9) - (4/8 (0)^8 + 9/9 (0)^9)
Simplifying further:
∫(C) z dx + x dy + y dz = (1/2 + 9/9) - (0 + 0)
The final result is:
∫(C) z dx + x dy + y dz = 1/2 + 1 = 3/2