Created on 2024-06-20Asked by Sebastian Young (Solvelet student)
Verify Stokes' Theorem for the vector field F=(y,−x,0) over the surface S which is the part of the plane z=0 inside the circle x2+y2=1.
Solution
To verify Stokes' Theorem for the vector field F=(y,−x,0) over the surface S which is the part of the plane z=0 inside the circle x2+y2=1: 1. **Compute the curl of F:** ∇×F=(∂y∂(0)−∂z∂(−x),∂z∂(y)−∂x∂(0),∂y∂(−x)−∂x∂(y))=(0,0,−2). 2. **Surface integral of ∇×F:** ∬S(∇×F)⋅dS=∬S(−2)⋅(0,0,1)dS=−2∬SdS=−2π(1)2=−2π. 3. **Line integral of F around the boundary C:** Parameterize C using x=cost,y=sint,0≤t≤2π: ∮CF⋅dr=∮C(ydx−xdy)=∫02π(sint⋅−sint−cost⋅cost)dt=∫02π(−sin2t−cos2t)dt=∫02π(−1)dt=−2π. 4. **Verification:** Both integrals are equal, confirming Stokes' Theorem: ∬S(∇×F)⋅dS=∮CF⋅dr=−2π.Solved on Solvelet with Basic AI Model
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DefinitionStokes Theorem-states that the surface integral of the curl of a vector field over a surface \(S\) is the same as the line integral of the vector field over the boundary curve \(\partial S\). In other words, the theorem states that \[ \int_{S}(\nabla \times \vec{F})\\cdot dS = \oint_{\partial S}\vec{F}\\cdot dr \] Example: This theorem simplifies complex surface integrals by replacing them with easier line integrals.
