A vector identity and surface integral

[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 <br /> \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)<br /> and \nabla \times \left( {f\mathop F\limits^ \to } \right) = f\nabla \times \mathop F\limits^ \to + \nabla f \times \mathop F\limits^ \to.

<br /> \nabla \bullet \left( {f\nabla g \times \nabla h} \right)<br />

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

<br /> = \nabla h \bullet \left( {\nabla \times f\nabla g} \right)<br />

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

<br /> = \nabla h \bullet \left( {\nabla f \times \nabla g} \right)<br /> 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.

 

[d_leet]

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

 

[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.

 

[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.

 

[d_leet]

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.

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.

 

[Benny]

Oh ok...I forgot about that.

 

[abszero]

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.

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.

 

[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.

 

[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)

 

[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.

 

[marlon]

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)

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