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Verify Stokes' theorem F=xi+z²j+y²k

Short answer

curl F = (2y−2z)i. On plane n=(i+j+k)/√3, dS=√3 dx dy, (curl F)·n dS = (2y−2z)dxdy = (2x+4y−2)dxdy. Surface integral over triangle x,y≥0, x+y≤1 gives 0. Line integral over boundary is also 0. Stokes' theorem verified.

Verify Stokes' theorem for F = x i + z² j + y² k over the plane surface x + y + z = 1 lying in the first octant.

Stokes' theorem: ∮_C F·dr = ∬_S (curl F)·n dS. Here F = x i + z² j + y² k and S is the triangle on x + y + z = 1 with vertices A(1,0,0), B(0,1,0), C(0,0,1).

Surface integral

curl F = (∂R/∂y − ∂Q/∂z) i + (∂P/∂z − ∂R/∂x) j + (∂Q/∂x − ∂P/∂y) k
       = (2y − 2z) i
n = (i + j + k)/√3,  dS = √3 dx dy
(curl F)·n dS = (2y − 2z) dx dy = (2y − 2 + 2x + 2y) dx dy = (2x + 4y − 2) dx dy
∬ over triangle x, y ≥ 0, x + y ≤ 1:
 = 2(1/6) + 4(1/6) − 2(1/2) = 1 − 1 = 0

Line integral (A → B → C → A)

F·dr = x dx + z² dy + y² dz.

AB: z = 0, y = 1 − x, x: 1→0:   ∫ x dx = −1/2          (other terms 0)
BC: x = 0, y = 1 − z, z: 0→1:   ∫ z² dy + y² dz
                                = ∫ (−z²) dz + (1 − z)² dz = −1/3 + 1/3 = 0
CA: y = 0, z = 1 − x, x: 0→1:   ∫ x dx = 1/2           (other terms 0)
Total = −1/2 + 0 + 1/2 = 0

Both sides are 0, so Stokes' theorem is verified.

✓ Answer checked by Claude Opus 5.5 (high effort)

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