1. I've just done a problem on parameterized volumes:

"Find a parameterization for the volume which lies below the cone z=r, above the x-y plane, and inside the cylinder x<sup>2</sup>+y<sup>2</sup>=1"

What I did was:

Ψ(r,θ)=(rcosθ,rsinθ,r)

0<θ<2π, 0<r<1 (<is less than or equal to)

Ψ(r,θ,t)=(rcosθ,rsinθ,tr)

0<θ<2π, 0<r<1, 0<t<1

I'm wondering why is it necessary to parameterize with t, since r is already between 0 and 1?

For a parameterized volume in R<sup>3</sup>, will you always have three parameters, as opposed to having two to represent a surface?...you might even be able to parameterize the volume with one parameter thoguh might you?

Does parameterizing a volume give any easier way to calculating the actual value of the volume (say r is in metres)? For this question it would be pretty easy to calculate the volume normally, but for more complicated parameterized volumes perhaps there is a method that can be used to speed things up/make things easier?...just something I was wondering about.

Thansk alot  2.

3. You'll never be able to "nicely" parameterize a volume or even an area with only one parameter, where by "nice" I mean functions which are, say, piecewise differentiable. Space-filling curves do exist, but they're really nasty (and also cool). In general, you need n variables to parameterize an n-dimensional "thing".

With your example, note that the region your parameterization describes is only the TOP "face" of the desired volume. That's why t is introduced--to get the points lying below the top.

You're right that parameterizations can help you calculate volumes more easily--if the volume you're looking at is defined via neither well-known geometric shapes nor functions z = f(x,y), then parameterizations are the way to go.  4. Thanks, I suppose there is no point parameterizing with one paramter then, and having a complicated function, when 3 parameters will do it much more simply. Is that what you're saying?

I understand the problem now, but I'm struggling a bit with this parameterizations section in general. I've done some more problems, and seem to be getting close to the answers, but no cigar. :?

Another problem is:

Describe the vloume given by:

Ψ(p,θ,φ)=(psinφcosθ, psinφsinθ,pcosφ)

1<p<2, 0<θ<2π, 0<φ<π/2

I put it is 1/4 of a sphere of radius 2, with a 1/4 sphere of radius one cut out. They put it is "the region between spheres of radii 1 and 2, and above the x-y plane". But surely it can't be the whole area above the x-y plane, if φ only goes to π/2? I'm wondering if it's another misprint on their part, cause I'm sure what I've written is right.

Also, How exactly would you calculate the volume from the parameterzation then? A volume integral?

Also, can you recommend any good graphing programs for the PC, where I can do 2D, 3D, and parameterized curves in cylindrical and spherical co-ordinates? I think it would make this section alot more rewarding for me. I used to use "autograph" which was really good, cause it had curve fitting and stuff too, and calculating functions from input data, which is a really useful thing....I got it free from school, lost it, and couldn't find anywhere to get it again. I recently downloaded "Graphcalc" for free, but the parameterized functions didn't seem too impressive, or extensive in what you could enter. I'm also not sure if they have curve fitting and stuff.

Thanks alot  5. Originally Posted by bit4bit
Thanks, I suppose there is no point parameterizing with one paramter then, and having a complicated function, when 3 parameters will do it much more simply. Is that what you're saying?
Basically, yes, but the emphasis shouldn't be on the fact that 3 parameters are easier than 1 parameter. The emphasis should be that it would be next to impossible to work with the functions required for using 1 parameter. Nature forces your hand, you have to work with 3 parameters to parameterize a volume.

Another problem is:

Describe the vloume given by:

Ψ(p,θ,φ)=(psinφcosθ, psinφsinθ,pcosφ)

1<p<2, 0<θ<2π, 0<φ<π/2

I put it is 1/4 of a sphere of radius 2, with a 1/4 sphere of radius one cut out. They put it is "the region between spheres of radii 1 and 2, and above the x-y plane". But surely it can't be the whole area above the x-y plane, if φ only goes to π/2? I'm wondering if it's another misprint on their part, cause I'm sure what I've written is right.
I see what you're saying, but you're ignoring the fact that θ goes up to 2π. If you let φ go up to π, you'd actually be covering the region described by the book twice. If you really want φ to go up to π, you'd have to restrict θ to the interval [0,π] to prevent covering the volume twice.

Also, How exactly would you calculate the volume from the parameterzation then? A volume integral?
So a parameterization is basically a nice change of variables. So you can start with a volume integral representing the volume of your solid, and then you use the parameterization to change variables. This changes the region of integration to the domain for your parameterization (notice that generally your domain is very nice when you parameterize!). But when you change variables, you have to account for the change in differentials. The tool for doing this for multiple variables is called the Jacobian. It's "the determinant of the matrix of partial derivatives of the parameterization". For details, check:

http://en.wikipedia.org/wiki/Jacobian
http://en.wikipedia.org/wiki/Substitution_rule

The upshot is that now you have an integral over a nice domain. The Jacobian will introduce some functions that you now have to worry about integrating, but hopefully the domain and parameterization functions are so nice that this shouldn't be too hard.

Also, can you recommend any good graphing programs for the PC, where I can do 2D, 3D, and parameterized curves in cylindrical and spherical co-ordinates?
Unfortunately, I can't--I don't use such programs. Sorry.  6. Thanks.

I see what you're saying, but you're ignoring the fact that θ goes up to 2π. If you let φ go up to π, you'd actually be covering the region described by the book twice. If you really want φ to go up to π, you'd have to restrict θ to the interval [0,π] to prevent covering the volume twice.
Oh yeh, I forgot about θ, but how can the shape even be defined in the area where φ is not in the interval [0,π/2], even when θ goes all the way to 2π? I can't get my head round these problems, I just keep trying to think about normal functions like f(x,y).

So a parameterization is basically a nice change of variables. So you can start with a volume integral representing the volume of your solid, and then you use the parameterization to change variables. This changes the region of integration to the domain for your parameterization (notice that generally your domain is very nice when you parameterize!). But when you change variables, you have to account for the change in differentials. The tool for doing this for multiple variables is called the Jacobian. It's "the determinant of the matrix of partial derivatives of the parameterization". For details, check:

http://en.wikipedia.org/wiki/Jacobian
http://en.wikipedia.org/wiki/Substitution_rule

The upshot is that now you have an integral over a nice domain. The Jacobian will introduce some functions that you now have to worry about integrating, but hopefully the domain and parameterization functions are so nice that this shouldn't be too hard.
Thanks, I've scanned ahead in my book, and the Jacobian is covered a little later on, so I'll tackle that when it comes. It's is in the section about 'calculus with parameterizations', but I think it'd be best if I understand the parameterizations first.   Bookmarks
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