If F=x²yi+y²zj+xyzk find Div F
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Answer:
F = z cos(e
y
2
)i + x
√
z
2 + 1 j + e
2y
sin 3x k.
div F = ∇ · F
=
∂
∂x z cos(e
y
2
) + ∂
∂y x
√
z
2 + 1 +
∂
∂z e
2y
sin 3x
= 0 + 0 + 0 = 0
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