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The divergence of the vector field x2i + 2y3j + z4k at x =1 , y= 2, z = 3 is
Correct answer is '134'. Can you explain this answer?
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The divergence of the vector field x2i + 2y3j + z4k at x =1 , y= 2, z...


Given Vector Field:
- F = x^2i + 2y^3j + z^4k

Calculation of Divergence:
- The divergence of a vector field F = Pi + Qj + Rk is given by div(F) = ∂P/∂x + ∂Q/∂y + ∂R/∂z.

Substitute the given values:
- P = x^2, Q = 2y^3, R = z^4
- ∂P/∂x = 2x, ∂Q/∂y = 6y^2, ∂R/∂z = 4z^3

Calculate the divergence at x=1, y=2, z=3:
- ∂P/∂x = 2(1) = 2
- ∂Q/∂y = 6(2)^2 = 24
- ∂R/∂z = 4(3)^3 = 108
- div(F) = 2 + 24 + 108 = 134

Therefore, the divergence of the given vector field at x=1, y=2, z=3 is 134.
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The divergence of the vector field x2i + 2y3j + z4k at x =1 , y= 2, z = 3 isCorrect answer is '134'. Can you explain this answer?
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