line integrals through scalar field

May 2010
7
1
I am familiar with the idea of a LINE INTEGRAL of a SCALAR QUANTITY, defined in (say) three-dimensional space, with respect to the arc length;
and this generally also results in a scalar quantity.

Stroud's Advanced Engineering Mathematics, in Programme 18, evaluates
this to a VECTOR - which I have not seen elsewhere - and gives no
indication as to what this really means.

I THINK the length of this vector is the quantity which would result
from the integration w.r.t. the arc. Stroud simply leaves the vector
as "the answer". I would be grateful for confirmation that another step is
needed.

Does the vector have an interpretation or is it just an intermediate result?
 
May 2010
274
67
Los Angeles, California
The line integral of a scalar field will be a scalar, not a vector. Could you give more mathematical details of what Stroud is saying. Type Stroud's definition which leads to a vector quantity. Maybe you're confusing what he's saying with the line integral of a vector field (which will still be a scalar).
 
May 2010
7
1
Stroud example

Thank you for taking an interest.

No, Stroud is not talking about the line integral of a vector field - which
he deals with later - nor the integral of a dot product.

Stroud has the following Example 1:

If V=x (y^2) z evaluate integral of Vdr (dr is a vector) along the curve

C having parametric equations x = 3u, y = 2u^2, z = u^3 between

A(0,0,0) and B(3,2,1).

He writes then V = 12 (u^8) and dr = i *3 du + j *4u du + k* 3u^2 du
(i,j,k being unit vectors)

and integrates 12 u^8 dr (expressed as above in terms of du) from u=0 to u=1 to get

4*i + (24/5)*j + (36/11)*k

which is left as "the answer".

I note that the length of this vector is just over 7 and I THINK that is also
the value if one integrates V along the arc. I tried it in MATLAB but it seemed unable to do it (!) and I do not have access at the moment to my preferred Maple.
which I know much better.
 
May 2010
274
67
Los Angeles, California
OK, I think I know what Stroud means. Some where in Programme 18 there should be a definition resembling:

\(\displaystyle \int_Cf(x,y,z)d\,{\bf r}=\int_Cf(x,y,z)dx\,{\bf i}+\int_Cf(x,y,z)dy\,{\bf j}+\int_Cf(x,y,z)dz\,{\bf k}\).

You then evaluate the components as standard line integrals. It's not usual to define "vector-valued" line integrals but Stroud has a few unusual definitons. For example, he talks about volume integrals of vector fields.
 
Last edited:
May 2010
7
1
Stroud

Yes, this is the definition.

I think it is just a different way of CALCULATING that standard
integral which generally uses arc-lengths, but in that case Stroud should have gone the extra step and said that the length of the vector is the ultimate objective.

I still wonder if the vector itself means anything.

I can see WHY another way of calculating the arc-length integral
can be handy. The arc-length integrand in this case is

12*u^8*Sqrt(9+16*u^2+9*u^4) (from u=0 to u=1)

which is impossible to do by hand. I put it into the on-line (indefinite) integrator at Wolfram's Mathematica and the answer was NUTS -
a page of elliptic integrals and imaginary numbers.

Matlab also integrated the (definite) integral to a complex number,
which is weird as the integrand is everywhere a simple real number.
Perhaps I should stipulate that the square root is positive, although
that is the convention.

Matlab evaluated the slightly simpler (and always larger) integrand

12*u^8*Sqrt(9+25*u^2)

to 7.226 while the length of the vector in Stroud is 7.05 - so I am
fairly happy now.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Thank you for taking an interest.

No, Stroud is not talking about the line integral of a vector field - which
he deals with later - nor the integral of a dot product.

Stroud has the following Example 1:

If V=x (y^2) z evaluate integral of Vdr (dr is a vector) along the curve
This V is a scalar, not a vector.

C having parametric equations x = 3u, y = 2u^2, z = u^3 between

A(0,0,0) and B(3,2,1).

He writes then V = 12 (u^8) and dr = i *3 du + j *4u du + k* 3u^2 du
(i,j,k being unit vectors)
Okay, because V is a scalar, Vdr is a vector which he integrates "component-wise".

and integrates 12 u^8 dr (expressed as above in terms of du) from u=0 to u=1 to get

4*i + (24/5)*j + (36/11)*k

which is left as "the answer".

I note that the length of this vector is just over 7 and I THINK that is also
the value if one integrates V along the arc. I tried it in MATLAB but it seemed unable to do it (!) and I do not have access at the moment to my preferred Maple.
which I know much better.
There are really three different kinds of integrals here.

The one you are referring to initially, "LINE INTEGRAL of a SCALAR QUANTITY, defined in (say) three-dimensional space, with respect to the arc length" is \(\displaystyle \int f(x,y,z)|d\vec{r}(x,y,z)|\). It is the integral of the product of two scalars and so is a scalar.

The integral of a vector function along a curve, \(\displaystyle \int \vec{f}(x,y,z)\cdot d\vec{r(x,y,z)}\) is the integral of a dot product. It is also the integral of a scalar and so is a scalar.

The integral here is the integral of a scalar along a curve (NOT "with respect to the arc length") \int f(x,y,z)d\vec{r}[/itex] and so is a vector.
 
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May 2010
7
1
To HallsOfIvy

Thank you; but is the length of that vector (which is "the answer"), in the third type of integral, EQUAL to the scalar result from arc-integration of V, in the first type of integral?

And does the vector answer actually have an interpretation or meaning?
 
May 2010
7
1
Stroud

I have now calculated with Maple 13 (beats Matlab and Mathematica, but
they were old versions), and the Length of the Vector IS INDEED the scalar arc-length integral.

The only question left is whether the Vector has a useful interpretation.