Help with change of variables please

dextercioby

Yes,so what...?

Daniel.

hypermorphism

Basically, what are you talking about ?

dextercioby

I'm just waiting to see what trick you pull to convert the square [0,1]*[0,1] in itself by a change of variable...

Could you come up with a drawing...?

Daniel.

ninjacookies

I'm sort of confused...I thought the Jacobian was a matrix computation of derivatives, and that [tex]rdrd\theta[/tex] was simply a representation of dA, or are the two synonamous? I mean, would I be able to show that the jacobian is [tex]rdrd\theta[/tex] through a derivative matrix?

thanks again!

hypermorphism

Hi ninja,

There are two similar things mathematicians mean when they talk about the Jacobian. The Jacobian used in modern differential geometry is a (IMO) simpler and more algebraic formulation of the Jacobian as used in multivariable calculus, where the Jacobian is just a scalar. The two yield equivalent expressions, but as I find the DG expression more intuitive (and easier to write than a matrix!), I use that expression instead. Your matrix will yield only r, while the algebra of differential forms allows DG to include drdt.

hypermorphism

Sure. As soon as I'm done taking out the garbage. Fun to be at home.

ninjacookies

OK, I'm an idiot. I forgot my math book and I don't have the formula for integrals regarding csc and sec for powers larger than 2....could you please help?

I'm trying to take the integrals of [tex]\csc^3\theta d\theta[/tex] and [tex]\sec^3\theta d\theta[/tex]


I'm kinda in a rush as I need to turn this in within an hour and a half

thanks again!

hypermorphism

Right. So here is
The Story of Polar Coordinates and The Unit Square!
The closed unit square is defined as the subset of R^2 [0,1]x[0,1]. As the boundary of the closed square has 2-dimensional measure 0, we will refer to both the closed and open unit squares as simply the unit square. We will assume that Cartesian coordinates are well-known. Polar coordinates are related to Cartesian coordinates (x,y) by the transformation T(x,y) = ([tex]\sqrt{x^2+y^2}[/tex], arctan(y/x)) = (r, t) wherever the transformation is well-defined and the continuous extension elsewhere.
Note that the lines that define the unit square in Cartesian coordinates are x=1, x=0 and y=1, y=0. We want to split the unit square along its diagonal thusly:
Unit square
because of the way polar coordinates vary. The diagonal represents a fundamental break in the behavior of whatever overlying function defines the upper bounds of the unit square, so unless the equations tell us otherwise, we will assume we need a piecewise cover broken at the diagonal. In Cartesian coordinates, our regions (triangles) are then easily translated.
We transform the lower triangle first. The line x=1 is the set of points {(1, y)} which transforms to the set of points {([tex]\sqrt{1+y^2}[/tex], arctan(y))} in (r,t) coordinates. Thus y = tan(t) and r = [tex]\sqrt{1 + tan^2\theta}[/tex] = sec(t) is the equation of the line represented in Cartesian coordinates by x=1. Thus the lower triangle in the diagram is bounded by the polar equations r=0 to r=sec(t) where t varies from 0 to pi/4. In order to see this, work backwards from the bounds and see whether you end up with the triangle pictured.
The line y=1 is similarly r=csc(t). The further bounds for the upper triangle are trivial and are already in the thread. Of course, I didn't do all this formal stuff; I just noticed the geometry with respect to sweeping r and related it to polar coordinates. You can easily graph the functions r=csc(t), r=sec(t), t=0, and t=pi/2 to see the unit square they bound.

Attached Images

unitsquare.png (2.0 KB, 19 views)

ninjacookies

Wow, that was an excellent explanation. I can't thank you enough! :)


However, the abtiderivative of [tex]csc^3\theta[/tex] is [tex]csc\theta(cot^2\theta +1)[/tex] (if I'm correct) for t goes from pi/4 to pi/2, isn't [tex]cot^3\theta[/tex] undefined for theta = pi/2 ? Since the cotangent is just 1/tan...and tangent is undefined for tan (pi/2). ? So I'm confused how I would continue on this calculation?

I got [tex]\sqrt{2} /3 [/tex] for the lower triangle...now I'm stuck on the upper triangle.

Any help would be appreciated...thanks!!!

hypermorphism

Use the identity cos^2(t) + sin^2(t) = 1 to derive 1 + tan^2(t) = sec^2(t) and remember d(tan(t)) = sec^2(t). Unfortunately, there's no fast way to integrate this other than application of integration by parts and using trig identities. The answer is a pretty ugly bunch of natural logs, but it works. You may do well to find a more convenient bunch of coordinates, if possible.

PS. the antiderivative of csc^3(t) is unfortunately not that simple. What are the steps you took to get there ?

ninjacookies

I kinda had an itch the antiderivates weren't that easy to compute :( Now I only have 20 minutes left... :( He said we could use a calculator to compute the integrals instead of doing all the computation by hand, but unfortunately I have a very outdated ti-85 and not one of those new shiny ti-92's or whatever. I'm so sorry to ask this of you, but since you already said you found the answer for both triangles, could you please let me know what you got? It's times like this I wish I had the extra 100 plus dollars for a new calculator, but unfortunately I don't.

I cannot thank you enough...I owe you big time. :)

hypermorphism

Heya ninja,
Wolfram Research provides an integrator online as a sample of their Mathematica Suite. Type in "(Sec[x])^3" to get an unwieldy symbolic expression for the integral and similarly for Csc[x].
My numerical integrator (the free GraphCalc) gives ~1.147/3 for the lower triangle and ~1.147/3 for the upper triangle (makes sense considering the symmetry of the surface).

Edit:Forgot to divide by 3.

heman

That was quite impressive Hyper...But does it really matters to calculate the other one too...
It seems obvious that both of them are equal and if u are calculating the above one with cosec ...what is the problem calculating with sec..
i am not getting the exact feel.Pls help...

hypermorphism

Hi heman,
You're right. Since the surface {(r, t, r)} (polar/cylindrical coords) is symmetric with respect to t, it's sufficient to compute the integral above one of the triangles and multiply by 2. For surfaces without symmetry, we don't inherit that convenience, though.

heman

that i realize Hyper...But why did u write the Cosec For the above triangle..What is the Problem with Sec...Does the eqn of the line y=1 was to be written or how the r is varying ....Pls Help

hypermorphism

Hi heman,
If you look at the square, you will see that for the upper triangle, if we measure by the angle the radius makes with the line t=pi/2 (the y-axis), we get the relationship cos(t₀) = 1/r. t₀ is related to the t of polar coordinates by t₀ = pi/2 - t, so we have 1/r = cos(pi/2 - t) = sin(t), so r = 1/sin(t) = csc(t).