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From: JEMebius on 3 Jul 2010 23:01
Nando wrote:
> Hello!
>
> I have parametrized an ellipse in order to calculate its area by an
> integral. In the end I will explain the whole problem (this is a sort
> of homework assignment).
>
> The ellipse is defined in R^3, z= 0; and x^2 /20 + y^2 /30 = 1
>
> my parametric function is:
> (let "p" stand for ro, and "o" for theta)
>
> g: ] 0, 1 [ x ] 0, 2 pi[ -> R^3
>
> g ( p, o) = sqr(20) p cos o, sqr (30) p sen o, 0 ) (the last one is a
> zero)
>
> so I have a double integral
>
> int from 0 to 1 and int from 0 to 2 pi
>
> Then, I need to put the Jacobian determinant in there.
>
> I find the derivative of my parametrization function:
>
> sqr(20) cos o -sqr(20 p sen o
> sqr(30) sen o sqr(30) cos o
> 0 0
>
> Since I have no idea how to find the determinant of this matrice I
> simple ignored the zeros and (amazingly) got the same answer as the
> teacher solutions: -20 sqr(6) pi (there is a -2 outside the double
> integral from previous calculations).
>
> But, if there were some different values there, instead of zero, I do
> not know how I could find the Jacobian determinant!
>
> Anyone wants to offer some advice?
> (another tricky situation is when I parametrize a simple curve in R^3,
> and get a column matrix as derivative (gradiant). How do I find a
> determinant then?!)
>
>
>
> ADENDA
> There is, of course, the possibility that I am solving this problem so
> wrongly that, accidentally I got the right answer. And so, I'm going
> to state the whole question here.
>
> I am supposed to calculate the work done by a vectorial field in an
> ellipse curve (this is a calculus assignment, not physics).
> I am also supposed to use Stoke's Theorem.
>
> The curve is the one stated above.
> It's direction is clockwise when seen from point (0,0,10)
>
> The vector field is H(x, y, z) = (-y, x, 3)
>
>
>
> I was taught to do this the following way:
>
> Stokes theorem states that (over-simplification!) the integral of the
> rotational equals the integral of work (the integral that calculates
> work). Since I want to know the work, using Stokes, I should find the
> rotational.
>
> My rotational is <0, 0, 2>
> My exterior normal vector is <0, 0, -1>
>
> (and so the -2 appears!)
>
>
> Thanks in advance for any help!
> (I have found little to no information on determinants of non square
> matrices...)


General observation on Jacobian matrices
----------------------------------------

It is all multilinear algebra in the vicinity of a point P on an M-dimensional manifold in
an N-dimensional Euclidean space.
It makes indeed sense to speak of the determinants of a non-square Jacobian matrix.

For example: a 2-manifold in 4D space; parametric equation

(u, x, y, z) = (u(a, b), x(a, b), y(a, b), z(a, b));

the derivatives du/da, ..., dz/da; du/db, ..., dz/db make the Jacobian matrix.

There are six 2x2 determinants, which are readily interpreted as the components of a 4D
bivector. In the case of the tangent plane to a surface in 4D space this is a nonzero
simple bivector. It is nonzero by the definition of "parametrisation of a manifold" and it
is a simple bivector because it is just the wedge product of two 4D vectors.
The components of a simple 4D bivector B satisfy Pluecker's relation

B12.B34 + B13.B42 + B14.B23 = 0

Ciao: Johan E. Mebius
From: Nando on 10 Jul 2010 10:25
On 4 Jul, 04:01, JEMebius <jemeb...(a)xs4all.nl> wrote:
> Nando wrote:
> > Hello!
>
> > I have parametrized an ellipse in order to calculate its area by an
> > integral. In the end I will explain the whole problem (this is a sort
> > of homework assignment).
>
> > The ellipse is defined in R^3, z= 0; and x^2 /20 + y^2 /30 = 1
>
> > my parametric function is:
> > (let "p" stand for ro, and "o" for theta)
>
> > g: ] 0, 1 [ x ] 0, 2 pi[ -> R^3
>
> > g ( p, o) = sqr(20) p cos o, sqr (30) p sen o, 0 )  (the last one is a
> > zero)
>
> > so I have a double integral
>
> > int from 0 to 1 and int from 0 to 2 pi
>
> > Then, I need to put the Jacobian determinant in there.
>
> > I find the derivative of my parametrization function:
>
> > sqr(20) cos o                 -sqr(20 p sen o
> > sqr(30) sen o                sqr(30) cos o
> > 0                                          0
>
> > Since I have no idea how to find the determinant of this matrice I
> > simple ignored the zeros and (amazingly) got the same answer as the
> > teacher solutions: -20 sqr(6) pi (there is a -2 outside the double
> > integral from previous calculations).
>
> > But, if there were some different values there, instead of zero, I do
> > not know how I could find the Jacobian determinant!
>
> > Anyone wants to offer some advice?
> > (another tricky situation is when I parametrize a simple curve in R^3,
> > and get a column matrix as derivative (gradiant). How do I find a
> > determinant then?!)
>
> > ADENDA
> > There is, of course, the possibility that I am solving this problem so
> > wrongly that, accidentally I got the right answer. And so, I'm going
> > to state the whole question here.
>
> > I am supposed to calculate the work done by a vectorial field in an
> > ellipse curve (this is a calculus assignment, not physics).
> > I am also supposed to use Stoke's Theorem.
>
> > The curve is the one stated above.
> > It's direction is clockwise when seen from point (0,0,10)
>
> > The vector field is H(x, y, z) = (-y, x, 3)
>
> > I was taught to do this the following way:
>
> > Stokes theorem states that (over-simplification!) the integral of the
> > rotational equals the integral of work (the integral that calculates
> > work). Since I want to know the work, using Stokes, I should find the
> > rotational.
>
> > My rotational is <0, 0, 2>
> > My exterior normal vector is <0, 0, -1>
>
> > (and so the -2 appears!)
>
> > Thanks in advance for any help!
> > (I have found little to no information on determinants of non square
> > matrices...)
>
> General observation on Jacobian matrices
> ----------------------------------------
>
> It is all multilinear algebra in the vicinity of a point P on an M-dimensional manifold in
>   an N-dimensional Euclidean space.
> It makes indeed sense to speak of the determinants of a non-square Jacobian matrix.
>
> For example: a 2-manifold in 4D space; parametric equation
>
> (u, x, y, z) = (u(a, b), x(a, b), y(a, b), z(a, b));
>
> the derivatives du/da, ..., dz/da; du/db, ..., dz/db make the Jacobian matrix.
>
> There are six 2x2 determinants, which are readily interpreted as the components of a 4D
> bivector. In the case of the tangent plane to a surface in 4D space this is a nonzero
> simple bivector. It is nonzero by the definition of "parametrisation of a manifold" and it
> is a simple bivector because it is just the wedge product of two 4D vectors.
> The components of a simple 4D bivector B satisfy Pluecker's relation
>
> B12.B34 + B13.B42 + B14.B23 = 0
>
> Ciao: Johan E. Mebius

Thanks Johan E. Mebius. I appreciate your effort. But your explanation
is way to advanced for my knowledges. Anyway, I think I understood how
to solve the exercises and I already took my exam. Let's hope for the
bes.

Thank you.
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