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krati jain
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Asked: 2023-05-22T15:45:49+00:00 In:

382. If a function is described by F = (3x + z, y2 − sin x2z, xz + yex5), then the divergence theorem value in the region 0<x<1, 0<y<3 and 0<z<2 will be a) 13 b) 26 c) 39 d) 51

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382. If a function is described by F = (3x + z, y2 − sin x2z, xz + yex5), then the divergence theorem value in the region 0<x<1, 0<y<3 and 0<z<2 will be a) 13 b) 26 c) 39 d) 51
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    4 Answers

    1. Answer: c
      Explanation: Div (F) = 3 + 2y + x. By divergence theorem, the triple integral of Div F in
      the region is ∫∫∫ (3 + 2y + x) dx dy dz. On integrating from x = 0->1, y = 0->3 and z = 0-
      >2, we get 39 units.

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    3. Answer: c
      Explanation: Div (F) = 3 + 2y + x. By divergence theorem, the triple integral of Div F in
      the region is ∫∫∫ (3 + 2y + x) dx dy dz. On integrating from x = 0->1, y = 0->3 and z = 0-
      >2, we get 39 units.

      • 0

    4. Answer: c
      Explanation: Div (F) = 3 + 2y + x. By divergence theorem, the triple integral of Div F in
      the region is ∫∫∫ (3 + 2y + x) dx dy dz. On integrating from x = 0->1, y = 0->3 and z = 0-
      >2, we get 39 units.

      • 0

    6. Answer: c
      Explanation: Div (F) = 3 + 2y + x. By divergence theorem, the triple integral of Div F in
      the region is ∫∫∫ (3 + 2y + x) dx dy dz. On integrating from x = 0->1, y = 0->3 and z = 0-
      >2, we get 39 units.

      • 0

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