# Vector Operations?

1. Jan 18, 2013

### Erbil

1. The problem statement, all variables and given/known data

$$\overrightarrow {F}=3xz^{2}i+2xyj-x^{2}k$$ $$\phi =3x^{2}-yz$$ are given vector and scalar fields, respectively.

a) $$\quad \operatorname{div}\left( \operatorname{grad}\phi \operatorname{div}\overrightarrow {F}\right) =\quad?$$

b) $$\quad \operatorname{curl}\left( \phi F\right) =\quad?$$

c) $$\quad \operatorname{div}\left( \phi F\right) =\quad?$$

d) $$\quad \overrightarrow {\nabla }\cdot \left( \nabla \phi \times \overrightarrow {F}\right) =\quad?$$

e) $$\quad \nabla \cdot \left( \overrightarrow {F}\nabla \phi \right) =\quad?$$

I know the operations such as the Gradient, Divergence, Curl, and Laplacian. But I don't have an idea what can I do in this kind of problems?

2. Relevant equations
http://en.wikipedia.org/wiki/Vector_calculus_identities

3. The attempt at a solution

a)I found gradient of scalar field and divergence of vector field.Gradient of scalar field is a vector and divergence of vector fields is a scalar.So how can I take the divergence of this?
No idea for others.I think I have to use some formula for calculate these..But which? I don't wanna a solution just I want to understand logic.Please help me to figure this:) Then I will try to do it myself.

And also,I'd like to say that I'm sorry for my bad English.

2. Jan 18, 2013

### Staff: Mentor

Vector*Scalar (better: written as scalar*vector) is a vector again, and has a divergence.

Vector*Vector: Scalar product ("dot product")
Vector x Vector: Cross-product
Scalar*Vector or Vector*Scalar: Scalar multiplication of the vector
Scalar*Scalar: Just like multiplication with real numbers

You can evaluate these expressions step by step.

3. Jan 18, 2013

### Erbil

@mfb; Is there any another way,because there are a lot of process with step by step? Now,I'm trying step by step.I will tell the result,but I don't know where can I control it?

4. Jan 18, 2013

### Staff: Mentor

WolframAlpha and other computer algebra systems should be able to do that.
There are many steps, but they are all easy, and you can even re-use some.

For more complicated expressions, vector identities can be useful, but I think you cannot use them (in a meaningful way) here.

5. Jan 18, 2013

### Erbil

OK.Here is my expression.

= (18z^2-12+72x+2z-6z^2+2y-6yz) + ( 24-4z-3z^2-8y+2x-6xz) + (36xz+36x^2-9z^2-4y+2x-12xz-6yz-6xy)

= 36x^2-6xy-12yz+6y+18xz am I right?