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From: José Carlos Santos on 3 Jul 2010 03:07
On 03-07-2010 1:47, Nando wrote:

> 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
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