Volume of a Parametrised Surface

  • Thread starter Thread starter Tom31415926535
  • Start date Start date
  • Tags Tags
    Surface Volume
Tom31415926535
Messages
9
Reaction score
0

Homework Statement


Let C be the parametrised surface given by

Φ(t,θ)=(cosθ/cosht, sinθ/cosht,t−tanht), for 0≤t and 0≤θ<2π

Let V be the region in R3 between the plane z = 0 and the surface C.

Compute the volume of the region V .

Homework Equations

The Attempt at a Solution


I thought I needed to perform a change of variables; changing from x,y,z to theta,t,z.

I tried to find the Jacobean for this, but it came to zero. I'm pretty sure it was wrong in the first place, but I have no idea what to do otherwise. Your assistance would be greatly appreciated.
 
Physics news on Phys.org
It's a surface of revolution. To visualize what it looks like, figure out what cross sections of constant ##t## are.
 
Here's a picture if that helps you. In the picture ##0\le t \le 5##.
picture2.jpg
 

Attachments

  • picture2.jpg
    picture2.jpg
    11.5 KB · Views: 283
Thanks heaps! That’s very helpful :smile:

I’m a little uncertain about how to setup the integral to calculate the volume though. Am I correct in needing to perform a change of variables? If so, what am I doing incorrectly that produces a Jacobean of zero?

Thanks :smile:
 
Tom31415926535 said:
Thanks heaps! That’s very helpful :smile:

I’m a little uncertain about how to setup the integral to calculate the volume though. Am I correct in needing to perform a change of variables? If so, what am I doing incorrectly that produces a Jacobean of zero?

Thanks :smile:

You don't do that sort of a change of variables. You have only two parameters so you have a two dimensional surface. In the three dimensions ##x, y, z## that will have zero volume. That's why your Jacobean is zero. The problem is actually easier than that. As I said you can treat it as a solid of revolution. https://en.wikipedia.org/wiki/Solid_of_revolution
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top