Stokes' theorem converts the line integral round C into:

The surface integral of (curl F).n over a surface bounded by C

The surface integral of F.n over C

The double integral of curl F over the xy-plane only

Evaluate using Stokes' theorem

Short answer

By Stokes' theorem, curl of F = eˣ i + 2y j − k is zero. Hence the line integral over the closed curve is 0.

Evaluate by Stokes' theorem ∮_C (eˣ dx + 2y dy − dz) where C is the curve x² + y² = 4, z = 2.

Stokes' theorem: ∮_C F·dr = ∬_S (curl F)·n dS.

Here ∮_C (eˣ dx + 2y dy − dz) = ∮_C F·dr with F = eˣ i + 2y j − k.

curl F = | i    j    k  |
         | ∂x   ∂y   ∂z |
         | eˣ   2y   −1 |
       = (0 − 0) i − (0 − 0) j + (0 − 0) k = 0

Since curl F = 0 on any surface S spanning C (for example the disc x² + y² ≤ 4 in the plane z = 2),

∮_C F·dr = ∬_S 0 · n dS = 0

Direct check: on C, z = 2 so dz = 0. With x = 2cos t, y = 2sin t: ∮ eˣ dx = [eˣ] over one full loop = 0, ∮ 2y dy = [y²] = 0, ∮ (−dz) = 0.

Answer: ∮_C (eˣ dx + 2y dy − dz) = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)