Question

Curls and divergences * Calculate the curl and the divergence of each of the following vec- tor fields. If the curl turns out to be zero, try to discover a scalar function φ of which the vector field is the gradient.

(a) F=(x+y,−x+y,−2z);

(b) G=(2y,2x+3z,3y);

(c) H=(x2 −z2,2,2xz).

Answer 1

a)
`\\abla F=0, \\abla* F=(0,0,-2)`

b)
`\\abla F=0, \\abla* F=(0,0,0)`,
`f=2xy+3yz+C, C\in\mathbb{R}`.

c)
`\\abla F=4x, \\abla* F=(0,-4z,0)`

Explanation:

Remember, if F= <f,g,h> is a vector field and
`\\abla` is the operator
`\\abla=<(\partial)/(\partial x), (\partial)/(\partial y), (\partial)/(\partial z)>`

• the divergence of F is
  `\\abla F=<(\partial)/(\partial x), (\partial)/(\partial y), (\partial)/(\partial z)>\cdot F= (\partial f)/(\partial x)+(\partial g)/(\partial y) +(\partial h)/(\partial z)`
• the curl of F is
  `\\abla* F=det(\left[\begin{array}{ccc}i&j&k\\(\partial)/(\partial x)&(\partial)/(\partial y)&(\partial)/(\partial z)\\f&g&h\end{array}\right] )`

a)
`F=(x+y,-x+y,-2z)`

The divergence of F is

`\\abla F=(\partial)/(\partial x)(x+y)+(\partial)/(\partial y)(-x+y)+(\partial)/(\partial z) (-2z)=1+1-2=0`

The curl of F is

`\\abla* F=det(\left[\begin{array}{ccc}i&j&k\\(\partial)/(\partial x)&(\partial)/(\partial y)&(\partial)/(\partial z)\\x+y&-x+y&-2z\end{array}\right])\\=i((\partial)/(\partial y)(-2z)-(\partial)/(\partial z)(-x+y))-j((\partial)/(\partial x)(-2z)-(\partial)/(\partial z)(x+y))+k((\partial)/(\partial x)(-x+y)-(\partial)/(\partial y)(x+y))=0i-0j-2k=(0,0,-2)`

b)
`F=(2y,2x+3z,3y)`

The divergence of F is

`\\abla F=(\partial)/(\partial x)(2y)+(\partial)/(\partial y)(2x+3z)+(\partial)/(\partial z) (3y)=0+0+0=0`

The curl of F is

`\\abla* F=det(\left[\begin{array}{ccc}i&j&k\\(\partial)/(\partial x)&(\partial)/(\partial y)&(\partial)/(\partial z)\\2y&2x+3z&3y\end{array}\right])\\=i((\partial)/(\partial y)(3y)-(\partial)/(\partial z)(2x+3z))-j((\partial)/(\partial x)(3y)-(\partial)/(\partial z)(2y))+k((\partial)/(\partial x)(2x+-3z)-(\partial)/(\partial y)(2y))=0i-0j+0k=(0,0,0)`

Since the curl of F is 0 the we will try find f such that the gradient of f be F.

Since
`f_x=2y`,
`f=2xy+g(y,z)`.

Since
`f_y=2x+3z`,
`2x+3z=f_y=2x+g_y(y,z)\\3z=g_y`.

Since
`f_z=3y` and
`f_z=3y+h'(z)`, then
`h'(z)=0`. Thins means that
`h(z)=C, C\in\mathbb{R}`

Therefore,

`f=2xy+3yz+C, C\in\mathbb{R}`.

c)
`H=(x^2-z^2,2,2xz)`

The divergence of F is

`\\abla F=(\partial)/(\partial x)(x^2-z^2)+(\partial)/(\partial y)(2)+(\partial)/(\partial z) (2xz)=2x+0+2x=4x`

The curl of F is

`\\abla* F=det(\left[\begin{array}{ccc}i&j&k\\(\partial)/(\partial x)&(\partial)/(\partial y)&(\partial)/(\partial z)\\x^2-z^2&2&2xz\end{array}\right])\\=i((\partial)/(\partial y)(2xz)-(\partial)/(\partial z)(2))-j((\partial)/(\partial x)(2xz)-(\partial)/(\partial z)(x^2-z^2))+k((\partial)/(\partial x)(2)-(\partial)/(\partial y)(x^2-z^2))=0i-(2z-(-2z))j-0k=(0,-4z,0)`