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Evaluate a line integral F=zi+xj+yk

Short answer

Parametrize: dx=2t dt, dy=4t dt, dz=3t² dt. F·dr = z dx + x dy + y dz = (8t⁴+4t³+4t) dt. Integrate from t=1 to 2: ∫(8t⁴+4t³+4t)dt = [8t⁵/5 + t⁴ + 2t²]₁² = 353/5 = 70.6.

Evaluate ∫_C F·dr where F = z i + x j + y k and C is the arc of the curve x = t² + 1, y = 2t², z = t³ from t = 1 to t = 2.

Here F·dr = z dx + x dy + y dz, and the curve is x = t² + 1, y = 2t², z = t³, with t from 1 to 2.

dx = 2t dt,  dy = 4t dt,  dz = 3t² dt

z dx = t³ · 2t dt = 2t⁴ dt
x dy = (t² + 1)(4t) dt = (4t³ + 4t) dt
y dz = 2t² · 3t² dt = 6t⁴ dt

F·dr = (8t⁴ + 4t³ + 4t) dt

∫₁² (8t⁴ + 4t³ + 4t) dt = [8t⁵/5 + t⁴ + 2t²]₁²
 = (256/5 + 16 + 8) − (8/5 + 1 + 2)
 = 248/5 + 21
 = 353/5

Answer: ∫_C F·dr = 353/5 = 70.6.

✓ Answer checked by Claude Opus 5.5 (high effort)

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