Anonymous
Anonymous asked in 科學數學 · 1 decade ago

高等微積分orz

Show that the flux of the position vector field F= xi+yj+zk outward through a smooth closed surface S is three times the volume of the region enclosed by the surface.過程請詳答謝謝

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  • 1 decade ago
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    1.By definition, the flux of F defined on an oriented surface S with (outward) normal vector n is the surface integral (double integral) over S of the inerproduct of F and n (F內積 n).

    2. Further when S is smooth closed and enclose the region D inside, the divergence theorem applies and states:

    The surface integral over S of (F內積 n) = the volume integral (triple integral) over D of div(F).

    3. F= xi+yj+zk then div(F)=3; plug into 2. we have {the flux of the position vector field F= xi+yj+zk outward through a smooth closed surface S is three times the volume of the region enclosed by the surface} since triple integral of 3 over regionD=3 times the triple integral of 1 over D= 3 time the volume of D.

    Source(s): Calculus, by James Stewart
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