What is the divergence of vector field F(x,y,z) = (-x+y)i + (y+z)j + (-z+x)k?

[1MileCrash]

Homework Statement

F(x,y,z) = (-x+y)i + (y+z)j + (-z+x)k

Find divergence

Homework Equations

The Attempt at a Solution

The gradient is
-i + j + -k

Dotting that with F, I get

x - y + y + z + z - x
=
2z

My book lists the answer as -1. What the heck are they talking about? (they did not ask me to evaluate for any point)

 

[Dick]

Homework Statement

F(x,y,z) = (-x+y)i + (y+z)j + (-z+x)k

Find divergence

Homework Equations

The Attempt at a Solution

The gradient is
-i + j + -k

Dotting that with F, I get

x - y + y + z + z - x
=
2z

My book lists the answer as -1. What the heck are they talking about? (they did not ask me to evaluate for any point)

I think you are misunderstanding the definition of divergence. ∇.F doesn't mean grad(F).F. Look it up.

 

[1MileCrash]

I think you are misunderstanding the definition of divergence. ∇.F doesn't mean grad(F).F. Look it up.

I will, thanks!

 

[1MileCrash]

Is it correct to say that it's like taking grad F, then adding up the resulting components for a scalar?

 

[Dick]

Is it correct to say that it's like taking grad F, then adding up the resulting components for a scalar?

Yes, if F=(Fx,Fy,Fz) then the divergence of F is dFx/dx+dFy/dy+dFz/dz. It's a scalar.

 

[1MileCrash]

I see my misunderstanding now. The del operator is not the gradient of anything in particular. It's just (d/dx)i + (d/dy)j + (d/dz)k. Dot product that with F leads to the correct definition.

This actually clears up a lot of the past notation. Since del is not a gradient of anything in particular, when we say \nabla f, since f is a scalar being multiplied by some vector, del, the result is a vector, which is the gradient of f.

Cool. :)

 

[Dick]

I see my misunderstanding now. The del operator is not the gradient of anything in particular. It's just (d/dx)i + (d/dy)j + (d/dz)k. Dot product that with F leads to the correct definition.

This actually clears up a lot of the past notation. Since del is not a gradient of anything in particular, when we say \nabla f, since f is a scalar being multiplied by some vector, del, the result is a vector, which is the gradient of f.

Cool. :)

You've got it, I think. You just dotting the grad operator with the vector. The result is a scalar.