It actually comes from a theorem. If the curl of a vector field is zero then the vector field is equal to the negative of the gradient of the scalar potential.

\nabla \times \mathbf{F} = \mathbf{0}

\Rightarrow \mathbf{F} = - \nabla \phi

Therefore you have an integral which is independent of the line and so:

\mathbf{F} = \nabla W

\Rightarrow \int^b_a \mathbf{F} d\mathbf{x} = \int^b_a dW = \left[W \right]^b_a

and I worked W out to be

||\mathbf{x}||^4

. I know this is correct, and yes a and b are scalars. What I don't understand is how to find:

\left[||\mathbf{x}||^4\right]^b_a

although in the solution the next step just gives the answer as:

(b^4-a^4)

Reply 3

DFranklin

I am unconvinced what you've written is a full and accurate description of the question.

I assume you're supposed to be doing something like "work out the work done in moving from a point at radius a to a point at radius b".

In which case, when the radius is a, you don't know what x is, but you do know its norm is a. (That is,

||{\bf x}|| = a

). Similarly you know the norm is b at the other limit. So the change in

||x||^4

is b^4 - a^4 as desired.

Reply 4

adie_raz

Original post by DFranklin

||{\bf x}|| = a

). Similarly you know the norm is b at the other limit. So the change in

||x||^4

is b^4 - a^4 as desired.

The question is this:

Show that the line integral of

\mathbf{A}(\mathbf{x}) = 4||\mathbf{x}||^2\mathbf{x}

along any two non-identical paths C1 and C2, each joining any point on the sphere

x^2+y^2+z^2=a^2

to any point on the sphere

x^2+y^2+z^2=b^2, (b>a>0)

, yields the same value.

Calculate this value as a function of a and b.

What I have done is utilised the theorem above to find the answer to integral, but do not understand how you put the bounds into the answer to find the final answer of

b^4-a^4

.

What you are saying does make sense to me (with regard work done etc) but I do need explaining how in this context you can just put a and b into the norm. (Apologies for not typing the question before, I did not think that the context of the question would be relevant).

Reply 5

ztibor

Original post by adie_raz

The question is this:

Show that the line integral of

\mathbf{A}(\mathbf{x}) = 4||\mathbf{x}||^2\mathbf{x}

along any two non-identical paths C1 and C2, each joining any point on the sphere

x^2+y^2+z^2=a^2

to any point on the sphere

x^2+y^2+z^2=b^2, (b>a>0)

b^4-a^4

THe centre of both sphere is the origin. The integral is independent from the
path so choose a path along the axis x with parameter of t.
THat is let

\vec{x}=[t,0,0]

and t starts at a and ends at b.

\int^{x_2}_{x_1} \vec{A}(\vec{x})\cdot d\vec{x}

Parametrizing A

4t^2\cdot t+0+0=4t^3

Integrating dot product of vector A and vector dx

\int^b_a 4t^3 dt =[t^4]^b_a

Consider that the

||x||=\sqrt{t^2}

t^4=||x||^4

(edited 13 years ago)

Reply 6

DFranklin

Original post by adie_raz

What you are saying does make sense to me (with regard work done etc) but I do need explaining how in this context you can just put a and b into the norm. (Apologies for not typing the question before, I did not think that the context of the question would be relevant).

What I've written is a perfectly adequate explanation of why the norm at the lower limit is a and at the upper limit is b. I don't think I can explain it any better.

(The integral you've calculated is exactly equivalent to "the work done" - the only reason I used those terms was I thought it was more likely you'd been asked "to find the work done in moving from A to B", rather than "

\int {\bf F.dx}

along a path from A to B". But the two phrases in quotes are exactly equivalent).

Reply 7

adie_raz

Original post by DFranklin

\int {\bf F.dx}

along a path from A to B". But the two phrases in quotes are exactly equivalent).

I see, so basically you can put in a and b to the norm due to the context of the question and the way in which it is asked which would also be true if I was asked to find the work done by moving along a path from a to b.

Thanks for the clarification. :biggrin: