The divergence of x2coszi⃗ +ylogxj⃗ −yzk⃗ 𝑥2𝑐𝑜𝑠𝑧𝑖→+𝑦𝑙𝑜𝑔𝑥𝑗→−𝑦𝑧𝑘→ is 0
Question
The divergence of x2coszi⃗ +ylogxj⃗ −yzk⃗ 𝑥2𝑐𝑜𝑠𝑧𝑖→+𝑦𝑙𝑜𝑔𝑥𝑗→−𝑦𝑧𝑘→ is 0
Solution
The divergence of a vector field F = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k is given by the formula:
div F = ∂P/∂x + ∂Q/∂y + ∂R/∂z
In this case, the vector field is F = x^2cosz i + ylogx j - yz k. So, we have:
P(x,y,z) = x^2cosz Q(x,y,z) = ylogx R(x,y,z) = -yz
We need to compute the partial derivatives of P, Q, and R with respect to x, y, and z, respectively.
∂P/∂x = 2xcosz ∂Q/∂y = logx ∂R/∂z = -y
So, the divergence of F is:
div F = ∂P/∂x + ∂Q/∂y + ∂R/∂z = 2xcosz + logx - y
Therefore, the divergence of the given vector field is not zero, unless for specific values of x, y, and z that satisfy the equation 2xcosz + logx - y = 0.
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