# A vector identity and surface integral

1. Dec 23, 2005

### Benny

Hi, can someone give me some assistance with the following questions?

1. Let f(x,y,z), g(x,y,z) and h(x,y,z) be any C^2 scalar functions. Prove that $$\nabla \bullet \left( {f\nabla g \times \nabla h} \right) = \nabla f \bullet \left( {\nabla g \times \nabla h} \right)$$ .

2. Let S be the part of the ellipsoid 2x^2 + y^2 + (z-1)^2 = 5 for z <=0 and $$\mathop F\limits^ \to = \left( {e^{y + z} + 3y,xe^{y + z} ,\cos \left( {xyz} \right) + z^3 } \right)$$ .

Evaluate $$\int\limits_{}^{} {\int\limits_S^{} {\left( {\nabla \times \mathop F\limits^ \to } \right)} \bullet d\mathop S\limits^ \to }$$.

(Use the normal to the surface pointing downwards.)

My working:

1. I will use $$\nabla \bullet \left( {\mathop A\limits^ \to \times \mathop B\limits^ \to } \right) = \mathop B\limits^ \to \bullet \left( {\nabla \times \mathop A\limits^ \to } \right) - \mathop A\limits^ \to \left( {\nabla \times \mathop B\limits^ \to } \right)$$ and $$\nabla \times \left( {f\mathop F\limits^ \to } \right) = f\nabla \times \mathop F\limits^ \to + \nabla f \times \mathop F\limits^ \to$$.

$$\nabla \bullet \left( {f\nabla g \times \nabla h} \right)$$

$$= \nabla h \bullet \left( {\nabla \times f\nabla g} \right) - f\nabla g \bullet \left( {\nabla \times \nabla h} \right)$$...from the identies above .The second bracket is zero since curl(grad(h)) = 0 vector.

$$= \nabla h \bullet \left( {\nabla \times f\nabla g} \right)$$

$$= \nabla h \bullet \left( {f\nabla \times \nabla g + \nabla f \times \nabla g} \right)$$...using identities listed above

$$= \nabla h \bullet \left( {\nabla f \times \nabla g} \right)$$ since curl(grad(g)) = 0 vector.

This is as far as I get in the first question.

For the surface integral I calculated $$\nabla \times \mathop F\limits^ \to = \left( { - xz\sin \left( {xyz} \right) - xe^{y + z} ,e^{y + z} + yz\sin \left( {xyz} \right), - 3} \right)$$.

I would normally parameterise the surface to find a normal to the surface. I'm not sure how to that or if I need to do that in this question.

Any help would be good thanks.

Last edited: Dec 23, 2005
2. Dec 23, 2005

### d_leet

For question #2 have you learned Stokes's theorem yet, that might help.

3. Dec 23, 2005

### StatusX

For the first one, there is the vector identity:

$$\vec A \cdot (\vec B \times \vec C) = \vec B \cdot (\vec C \times \vec A) =\vec C \cdot (\vec A \times \vec B)$$

This operation is called the triple product. For the second, use Stoke's theorem to turn the surface integral into a contour integral.

4. Dec 24, 2005

### Benny

Thanks for the help guys.

The thing is, Stokes' theorem was what I thought about using but the question says something about using a normal so I figured that I need to evaluate it as a surface integral. In any case if I was to use Stoke's theorem, I would need to parameterise an ellipse. Perhaps x = acos(t) and y = bsin(t) where a and b are suitable constants? But if sub those into my expression for curl(F) it looks like it'll get quite messy even with z = 0 getting rid of the trig functions.

5. Dec 24, 2005

### d_leet

But you wouldn't put them into curl(F) you would plug them into F because stokes' theorm says that the line integral of F dotted with dr is the same as the surface integral of curl(F) blah blah blah.

6. Dec 24, 2005

### Benny

7. Dec 24, 2005

### abszero

It's worth noting that this identity is a little trickier when you start throwing $$\nabla$$ into the mix. However, there is an identity like it that you might find useful.

8. Dec 25, 2005

### Benny

When you 'del' a scalar function you get a vector so it seems reasonable to suggest that it works. However, I haven't seen/(can't recall) any complete derivations of the cross product so I'm not sure if it matters if the vector has variable components.

9. Dec 25, 2005

### abszero

The problem is that del is a differential operator, a one-form, not a vector in the traditional sense, so there's an issue with ordering and such. For instance, what is $$\mathbf{b} \cdot (\mathbf{c} \times \nabla)$$

10. Dec 25, 2005

### StatusX

That's true, but I was referring to:

$$\nabla f \cdot (\nabla g \times \nabla h )$$

as it appears in his problem, and these are all vectors.

11. Dec 25, 2005

### marlon

Well first apply cyclic permutation to get the nabla out of the brackets. Then nabla must work on both b and c, so you will get a sum of two parts. The clue always is to write this formula in a sum of terms in which each nabla operates on ONE vector using the vector identities in the first post of this thread.

Ofcourse it can happen that you have this
$$\nabla \cdot \vec{b}$$

or

$$\vec{b} \cdot \nabla$$

The first formula is easy, right ? Just calculate the dot product and you will get a scalar.

The second formula actually defines a new scalar operator

regards
marlon

Last edited: Dec 25, 2005