Jacobian determinant from non square Jacobiam matrix (and Stokes' Theorem)
Nando
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...)
José Carlos Santos
> 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
Since you are working with an ellipse located at the plane z = 0, why
don't you just work in R^2?
> 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
That's natural; the determinant is defined only for square matrices.
>
> 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).
That is not amazing; your teacher worked in R^2 and therefore got a
2 x 2 matrix, which is equal to the matrix formed by the first two lines
of yours.
> 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?!)
Again, the determinant is defined only for square matrices.
Best regards,
Jose Carlos Santos
Stephen Montgomery-Smith
> 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.
You are embedding a two dimensional surface into three dimensional
space. So you should really be computing the integral of the norm of
the vector normal to the surface.
Your vector normal to the surface is ...
> 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
the cross produce of the two columns of the matrix you wrote here. And
if you look at the cross product, it will be
(0, 0, det)
where det is the determinant of the top 2x2 matrix.
>
> 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!
You would only get non-zero values there if the surface is not parallel
to the xy-plane.
>
> 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?!)
Take the norm of the derivative, and integrate that. (Don't use the
word "gradient" in this context, as that usually means something else.)
Nando
Thanks Stephen Montgomery-Smith.
So, my follow up:
I had done some research before posting. I studied (and found stated
in several places) that determinants only exist for square matrices.
However I've found the Jacobian determinant sometimes defined as a
sort of cross product, and visiting Mathword (http://
mathworld.wolfram.com/Jacobian.html) I got the idea that you can in
fact have a Jacobian determinant of 2 functions and 3 variables (and
therefore, not squared). Although the explanation there is way to
condensed for me to get any meaning of it.
As for the integral of the norm of the vector normal to the surface,
that was not the way I learnt to do surface integrals! But I'll be
researching it and trying that method, which seems far simpler.
I learned to do the integral of the square root of the matrix
resultant of the product of the transpose Jacobian matrix times the
(original) Jacobian matrix. I even tried that way, but it was taking
me way to long to be effective (during an exam I mean).
The formula I found in Wikipedia resembles the formula I use to
calculate a Flux over a vector field... Wonder if I am on to something
here...
Thanks again for your time and help!
On 3 Jul, 16:21, Stephen Montgomery-Smith <step...@math.missouri.edu>
wrote:
JEMebius
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
Nando
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.