Surface integral of angle between fixed vector and the unit normalS is the smooth solid boundary of an object in R3, and v is its outward unit normal. l is fixed vector in R3. The angle between l and v is θ. Prove: ∬ S cos ⁡ ( θ ) d S = 0.

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Surface integral of angle between fixed vector and the unit normalS is the smooth solid boundary of an object in R3, and v is its outward unit normal. l is fixed vector in R3. The angle between l and v is   θ. Prove:       ∬    S    cos  ⁡  (  θ  )        d    S    =  0.

Kaeden Lara · Accepted Answer

Step 1Let             F      →        be some fixed vector (I hate to use the symbol l which is easily confused with 1).Then since F is constant,   ∇  ⋅            F      →        =  0 everywhere. And in particular, if V is the volume enclosed by the surface in question,       ∫    V    ∇  ⋅            F      →          d  V  =  0.Now look at   cos  ⁡  θ. Since at each point on the surface, with normal             n      ^      ,             F      →        ⋅            n      ^        =      |    F      |    cos  ⁡  θ we can say that   cos  ⁡  θ  =                              F          →                    ⋅                        n          ^                                    |            F              |            .Step 2Therefore       ∫    S    cos  ⁡  θ    d  S  =      ∫    S                                F          →                    ⋅                        n          ^                                    |            F              |                d  S  =      1                  |            F              |                  ∫    S              F      →        ⋅            n      ^          d  S.Now use Gauss&#039;s law (another name for that is the divergence theorem) which says       ∫    S              F      →        ⋅            n      ^          d  S  =      ∫    V    ∇  ⋅            F      →          d  V to get       ∫    S    cos  ⁡  θ    d  S  =      1                  |            F              |                  ∫    S              F      →        ⋅            n      ^          d  S  =      1                  |            F              |                  ∫    V    ∇  ⋅            F      →          d  V  =      1                  |            F              |              0  =  0.

MMDCCC50m · Accepted Answer

Step 1Since I is constant, we may assume it extends to a vector field defined on all of             R        n   (we don&#039;t need to restrict ourselves to   n  =  3). Also, since I is constant, we have                     (1)                    div        (        I        )        =        ∇        ⋅        I        =        0        ;             furthermore, if   I  ≠  0,Step 2                    (2)                    v        ⋅        I        =        ‖        I        ‖        cos        ⁡        θ        .            We now apply the divergence theorem to I over the &quot;object&quot;       O   bounded by S:                    (3)                              ‖          I          ‖                      ∫            S                    cos          ⁡          θ                    d          S          =                      ∫            S                    I          ⋅          v          d          S          =                      ∫                                          O                                              ∇          ⋅          I          d          V          =                      ∫                                          O                                              0          d          V          =          0          ,                     the desired result, provided   I  ≠  0.

S is the smooth solid boundary of an object in R3, and v is its outward unit normal. l is fixed vector in R3. The angle between l and v is theta. Prove: int int_S cos (theta)dS=0.

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