Divergence of the vector field V(x,y,z) = -(x cos xy + y)i + (y cos xy)j + (sin z² + x² + y²)k is

(A) 2z cos z²
(B) sin xy + 2z cos z²
(C) x sin xy - cos z
(D) none of these

Correct answer: (A) 2z cos z²

Solution

div V = d/dx[-(x cos xy + y)] + d/dy[y cos xy] + d/dz[sin z^2 + x^2 + y^2]. This gives (-cos xy + xy sin xy) + (cos xy - xy sin xy) + 2z cos z^2; the cos xy and xy sin xy terms cancel, leaving 2z cos z^2, which is option A.

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