Volume of a Parametrised Surface

In summary, The given parametrised surface C is defined by the function Φ(t,θ)=(cosθ/cosht, sinθ/cosht,t−tanht), with t ranging from 0 to infinity and θ ranging from 0 to 2π. The region V in R3 is bound by the plane z = 0 and the surface C. To compute the volume of V, a change of variables is not necessary as it can be treated as a surface of revolution. Thus, the Jacobean will be zero and the problem can be simplified to finding the volume of a solid of revolution.
  • #1
Tom31415926535
9
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.
 
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  • #2
It's a surface of revolution. To visualize what it looks like, figure out what cross sections of constant ##t## are.
 
  • #3
Here's a picture if that helps you. In the picture ##0\le t \le 5##.
picture2.jpg
 

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  • #4
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:
 
  • #5
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
 

Related to Volume of a Parametrised Surface

1. What is the definition of volume for a parametrised surface?

The volume of a parametrised surface is the measure of space contained within the surface. It is calculated by finding the triple integral of the function representing the surface.

2. How is the volume of a parametrised surface calculated?

The volume is calculated using a triple integral, where the bounds of integration are determined by the parametric equations that define the surface. The integrand is the function representing the surface.

3. Can the volume of a parametrised surface be negative?

No, the volume of a parametrised surface cannot be negative as it is a measure of space and cannot have a negative value. It is always a positive quantity.

4. Are there any specific parametric equations that are commonly used for calculating the volume of a surface?

Yes, there are several commonly used parametric equations for calculating the volume of a surface, such as cylindrical coordinates, spherical coordinates, and parametric equations for specific geometric shapes like cones and spheres.

5. How is the volume of a parametrised surface related to its curvature?

The volume of a parametrised surface is not directly related to its curvature. However, the curvature of the surface can affect the bounds of integration and the integrand used to calculate the volume. Generally, surfaces with higher curvatures will have smaller volumes compared to surfaces with lower curvatures.

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