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junaid ansari
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Asked: 2024-10-08T21:01:03+05:30 In:

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

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    2 Answers

    1. junaid ansari

      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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    2. Quearn
      Best Answer
      Quearnist

      To solve for the divergence of a vector field (F = (3x + z, y^2 – sin(x^2z), xz + ye^{x^5})) and then use the divergence theorem to find the value in the specified region, we need to follow these steps:

      1. Find the divergence of (F):

      The divergence of a vector field (F = (P, Q, R)) is given by:

      [ nabla cdot F = frac{partial P}{partial x} + frac{partial Q}{partial y} + frac{partial R}{partial z} ]

      For (F = (3x + z, y^2 – sin(x^2z), xz + ye^{x^5})), we have:

      – (P(x, y, z) = 3x + z)

      – (Q(x, y, z) = y^2 – sin(x^2z))

      – (R(x, y, z) = xz + ye^{x^5})

      So,

      [ frac{partial P}{partial x} = 3 ]

      [ frac{partial Q}{partial y} = 2y ]

      [ frac{partial R}{partial z} = x ]

      Therefore, the divergence of (F) is:

      [ nabla cdot F =

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