# line integrals through scalar field

#### nigel

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?

#### ojones

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).

#### nigel

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.

#### ojones

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.

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#### nigel

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
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.

• AllanCuz

#### nigel

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?

#### nigel

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.