Parameterizing surface in surface integral problem

In summary, the problem is to find the surface area of the paraboloid z=2x^2+2y^2 cut by the planes z=2 and z=8. The surface area formula is given as S= ∫∫ ||Ts×Tt|| ds dt. The parameterization of the surface is (s,t,2s^2+2t^2) and the curves formed by the planes in the (s,t) plane are circles of radius 1 and 2 respectively. To solve the problem, polar coordinates can be used and the next step would be to find the partial derivative vectors and their cross product for integration.
  • #1
tasveerk
24
0

Homework Statement


Find the area of the surface cut from the paraboloid z=2x2+2y2 by the planes z=2 and z=8.

Homework Equations


Surface area of S= ∫∫ ||Ts×Tt|| ds dt

The Attempt at a Solution


What I am really having trouble doing in this problem (and in general) is parameterizing the surface in terms of s and t. I think for this surface an accurate parameterization is (s,t,√(2s2+2t2)) but I am not sure. Also, I wouldn't know how the planes cutting the paraboloid would come into play.
 
Physics news on Phys.org
  • #2
Why did you put the square root into (s,t,√(2s^2+2t^2)) for a starting point? The planes will define the (s,t) region you will integrate over.
 
  • #3
Good question. I seem to have made an error. I was looking at the parameterization of a different paraboloid and must have gotten mixed up.
 
  • #4
tasveerk said:
Good question. I seem to have made an error. I was looking at the parameterization of a different paraboloid and must have gotten mixed up.

Ok, so just use (s,t,2s^2+2t^2). Now if z=2, then that means 2s^2+2t^2=2. What kind of curve is that in the (s,t) plane?
 
  • #5
A circle of radius 1. When z=8, the curve will be a circle of radius 2. Not sure where to go from here.
 
  • #6
tasveerk said:
A circle of radius 1. When z=8, the curve will be a circle of radius 2. Not sure where to go from here.

Ok again. So you are going integrate between those two circles. Suggests you might think about polar coordinates. I'm not sure why you are not sure where to go. The next thing would be to work on what to integrate, right? Try working on the partial derivative vectors and finding the cross product.
 

What is the purpose of parameterizing a surface in a surface integral problem?

The purpose of parameterizing a surface is to represent the surface using a set of parameters, such as u and v, which allows us to break down the surface into smaller, more manageable pieces. This makes it easier to calculate the surface integral, as we can use the parameters to define a region on the surface and perform the integration over that region.

How do you parameterize a surface?

To parameterize a surface, we need to find a set of equations that describe the surface in terms of two variables, usually u and v. These equations can be found by manipulating the original equation of the surface, or by using geometric or graphical methods. The choice of parameters is usually arbitrary, but it is important to choose parameters that cover the entire surface without any gaps or overlaps.

What is the difference between a surface integral and a regular integral?

A surface integral is used to calculate the flux, or flow, of a vector field through a surface, while a regular integral is used to find the area under a curve or the volume of a solid. In a surface integral, we are integrating over a two-dimensional surface, whereas in a regular integral, we are integrating over a one-dimensional curve or a three-dimensional solid.

What are some common applications of surface integrals?

Surface integrals have many applications in physics and engineering, such as calculating the electric flux through a surface, determining the work done by a force on a surface, or finding the mass of a curved object. They are also used in computer graphics to render three-dimensional objects and in fluid dynamics to calculate the flow of a fluid over a surface.

How do you evaluate a surface integral?

To evaluate a surface integral, we first need to parameterize the surface and define the region of integration. Then, we use the parameterized equations to find the limits of integration for u and v. Next, we calculate the partial derivatives of the parameterized equations with respect to u and v, which will give us the components of the surface normal vector. Finally, we plug these values into the surface integral formula and perform the integration to get the final result.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
986
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
775
Replies
1
Views
556
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
488
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
21
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top