Example 7-11. Find the divergence of the vector field given by
v =xyz eˆ 1 +yz^2 ˆe 2 +zxy^2 ˆe 3
Solution By definition
div v =∇·v =
(
∂
∂x ˆe^1 +
∂
∂y eˆ^2 +
∂
∂z ˆe^3
)
·
(
xyz ˆe 1 +yz^2 ˆe 2 +zxy^2 ˆe 3
)
div v =∇·v =yz +z^2 +xy^2
Example 7-12. Find the curl of the vector field v =xyz ˆe 1 +yz^2 eˆ 2 +zxy^2 ˆe 3
Solution By definition
curlv =∇× v =
∣∣
∣∣
∣∣
∣
ˆe 1 ˆe 2 eˆ 3
∂
∂x
∂
∂y
∂
∂z
xyz yz^2 zxy^2
∣∣
∣∣
∣∣
∣
=ˆe 1
∣∣
∣∣
∂
∂y
∂
∂z
yz^2 zxy^2
∣∣
∣∣−ˆe 2
∣∣
∣∣
∂
∂x
∂
∂z
xyz zxy^2
∣∣
∣∣+ˆe 3
∣∣
∣∣
∂
∂x
∂
∂y
xyz yz^2
∣∣
∣∣
curlv =∇× v = (2xyz − 2 yz)ˆe 1 −(zy^2 −xy )ˆe 2 −xz eˆ 3
Properties of the Gradient, Divergence and Curl
Let u=u(x, y, z)and v=v(x, y, z )denote scalar functions which are continuous
and differentiable everywhere and let A=A(x, y, z)and B =B(x, y, z )denote vector
functions which are continuous and differentiable everywhere. One can then verify
that the del or nabla operator has the following properties.
(i) grad (u+v) = grad u+ grad vor ∇(u+v) = ∇u+∇v
(ii) grad (uv ) = ugrad v+vgrad uor ∇(uv ) = u∇v+v∇u
(iii) grad f(u) = f′(u) grad uor ∇(f(u)) = f′(u)∇u
(iv) |grad u|=|∇ u|=
√(
∂u
∂x
) 2
+
(
∂u
∂y
) 2
+
(
∂u
∂z
) 2
(v) If a vector field is irrotational curlF = 0 , then it is derivable from a scalar
function by taking the gradient, then one can write F =F(x, y, z ) = grad u(x, y, z),
or F =∇u. The vector field F is called a conservative vector field. The function
ufrom which the vector field is derivable is called the scalar potential.
(vi) ∇·(A+B) = ∇·A+∇·B or div (A+B) = div A+ div B
(vii) ∇× (A+B) = ∇× A+∇× B or curl (A+B) = curl A+ curlB
(viii) ∇(uA) = (∇u)·A+u(∇·A)
(ix) ∇× (uA) = (∇u)×A+u(∇× A)
(x) ∇·(A×B) = B·(∇× A)−A·(∇× B)
(xi) ∇× (A×B) = (B·∇ )A−B(∇·A)−(A·∇ )B+A(∇·B)
