Question

Given the vector field F = zi + 2yzj + (x + y 2 )k.

(a) Find the divergence of the field.
(b) Find the curl of the field.
(c) Is the given field conservative? If it is, find a potential function.
(d) Compute Z C z dx + 2yz dy + (x + y 2 ) dz where C is the positively oriented curve y 2 + z 2 = 4, x = 5.
(e) Compute Z C z dx + 2yz dy + (x + y 2 ) dz where C consists of the three line segments: from (0, 0, 0) to (4, 0, 0), from (4, 0, 0) to (2, 3, 1), and from (2, 3, 1) to (1, 1, 1).

Answer 1

`\vec F(x,y,z)=z\,\vec\imath+2yz\,\vec\jmath+(x+y^2)\,\vec k`

a. The divergence of
`\vec F` is

`\\abla\cdot\vec F=(\partial(z))/(\partial x)+(\partial(2yz))/(\partial y)+(\partial(x+y^2))/(\partial z)`

`\boxed{\\abla\cdot\vec F=2z}`

b. The curl of
`\vec F` is

`\\abla*\vec F=\left((\partial(x+y^2))/(\partial y)-(\partial(2yz))/(\partial z)\right)\,\vec\imath+\left((\partial(z))/(\partial z)-(\partial(x+y^2))/(\partial x)\right)\,\vec\jmath+\left((\partial(2yz))/(\partial x)-(\partial(z))/(\partial y)\right)\,\vec k`

`\boxed{\\abla*\vec F=\vec0}`

c.
`\vec F` is conserviatve if there is a scalar function
`f` for which
`\vec F=\\abla f`. This means we would need

`(\partial f)/(\partial x)=z`

`(\partial f)/(\partial y)=2yz`

`(\partial f)/(\partial z)=x+y^2`

From these conditions we get

`f(x,y,z)=xz+g(y,z)`

`(\partial f)/(\partial y)=(\partial g)/(\partial y)=2yz\implies g(y,z)=y^2z+h(z)`

`f(x,y,z)=xz+y^2z+h(z)`

`(\partial f)/(\partial z)=x+y^2+(\mathrm dh)/(\mathrm dz)=x+y^2\implies(\mathrm dh)/(\mathrm dz)=0\implies h(z)=C`

So we do find a potential function
`f`,

`\boxed{f(x,y,z)=xz+y^2z+C}`

and
`\vec F` is indeed conservative.

d. Since
`\vec F` is conservative, and
`C` is closed circle, the integral of
`\vec F` along
`C` is 0.

e. Since
`\vec F` is conservative, its integral along
`C` depends only on the endpoints. In particular,

`\displaystyle\int_C\\abla f\cdot(\mathrm dx,\mathrm dy,\mathrm dz)=f(1,1,1)-f(0,0,0)=\boxed{2}`