What is an example of a surface integral using the method of projection?

In summary: Oh wow, I've been a very stupid person. The whole motivation of the problem was to find the surface area of a crisp of that shape - so when given the equation of such a surface, my mind decided to believe that the equation of the crisp was given in full by the equation - i.e the surface magically cuts off at the end of the crisp, but obviously only a finite section (the circle) is a model of the crisp....
  • #1
fayled
177
0
I'm a little unsure about an example of a surface integral I've come across, in which the method of projection is used.

The example finds the surface area of a hyperbolic paraboloid given by z=(x2-y2)/2R bounded by a cylindrical surface of radius a, such that x2+y2=<2. The first issue I'm having is in understanding the function of the cylinder - if we project the area element of the surface onto the xy plane then integrate this, surely the projection of the whole paraboloid onto this plane is not in the form of a circle, thus finding the integral of the projected area element would either involve limits too large or small, i.e we would get too much or too little of the surface.

If it helps the working is
S=∫dxdy√(1+x2/R2+y2/R2) where the latter square root is the projection factor between dS (on the surface) and dA (on the xy plane).
S=∫ (from zero to a) 2πrdr√(1+r2/R2). I'm also not sure about what is going on in this step - some kind of change of coordinates maybe?
Then let u=r2/R2 and solve for (2∏R2/3)[(1+a2/R2)3/2-1]

Thanks for any help in advance.
 
Physics news on Phys.org
  • #2
fayled said:
I'm a little unsure about an example of a surface integral I've come across, in which the method of projection is used.

The example finds the surface area of a hyperbolic paraboloid given by z=(x2-y2)/2R bounded by a cylindrical surface of radius a, such that x2+y2=<2. The first issue I'm having is in understanding the function of the cylinder - if we project the area element of the surface onto the xy plane then integrate this, surely the projection of the whole paraboloid onto this plane is not in the form of a circle, thus finding the integral of the projected area element would either involve limits too large or small, i.e we would get too much or too little of the surface.

If it helps the working is
S=∫dxdy√(1+x2/R2+y2/R2) where the latter square root is the projection factor between dS (on the surface) and dA (on the xy plane).
S=∫ (from zero to a) 2πrdr√(1+r2/R2). I'm also not sure about what is going on in this step - some kind of change of coordinates maybe?
Then let u=r2/R2 and solve for (2∏R2/3)[(1+a2/R2)3/2-1]

Thanks for any help in advance.

I'm not sure what you are dubious about. You aren't projecting the whole hyperboloid, you only projecting the part that's inside the cylinder. The step you aren't sure about is just changing the integral to polar coordinates.
 
  • #3
Dick said:
I'm not sure what you are dubious about. You aren't projecting the whole hyperboloid, you only projecting the part that's inside the cylinder. The step you aren't sure about is just changing the integral to polar coordinates.

Hmm how can we find the whole surface area if we aren't projecting the whole hyperboloid?
 
  • #4
fayled said:
Hmm how can we find the whole surface area if we aren't projecting the whole hyperboloid?

How can you find out the radius of a circle from knowing its diameter? geometrically if you know the shape you generally only need a cross section to imply the form of the object at least in simple 3d euclidean geometry, although of course there are certain rules, as in the rule of the length of a cone and so on. It gets a bit more awkward otherwise for shapes that are not uniform or at least bound by pi, but for example in calculus we know the shape of a circle is represented by pir^2 if we integrate that what do we get? Hence the rule 4/3pir^3 etc etc, they are probably essentially trying to get you to derive the general rules of a particular type of object. Hence trig identities etc. You can assume a lot from just knowing a radius in some defined cases as long as x or r is non 0.
 
Last edited:
  • #5
fayled said:
Hmm how can we find the whole surface area if we aren't projecting the whole hyperboloid?
How could we find "the whole surface area" if we were? The "whole hyperboloid" has infinite surface area. We must specify a finite subsection in order to get a finite surface area.
 
  • #6
HallsofIvy said:
How could we find "the whole surface area" if we were? The "whole hyperboloid" has infinite surface area. We must specify a finite subsection in order to get a finite surface area.


Oh wow, I've been a very stupid person. The whole motivation of the problem was to find the surface area of a crisp of that shape - so when given the equation of such a surface, my mind decided to believe that the equation of the crisp was given in full by the equation - i.e the surface magically cuts off at the end of the crisp, but obviously only a finite section (the circle) is a model of the crisp. Thanks!
 

1. What is a surface integral example?

A surface integral example is a mathematical problem that involves calculating the total value of a function over a surface. This can be represented as the double integral of the function over the surface.

2. How is a surface integral example different from a regular integral?

A regular integral calculates the area under a curve in two dimensions, whereas a surface integral calculates the value of a function over a surface in three dimensions. This means that a surface integral involves an extra dimension and is represented by a double integral instead of a single integral.

3. What is the formula for calculating a surface integral?

The formula for calculating a surface integral is given by ∫∫f(x,y) dS, where f(x,y) is the function being integrated and dS represents the infinitesimal element of surface area over which the function is being integrated.

4. What are some real-world applications of surface integral examples?

Surface integral examples are used in various fields such as physics, engineering, and computer graphics. They can be used to calculate the flux of a vector field through a surface, to find the mass of a 3D object, and to calculate surface area and volume of 3D shapes.

5. What are some common techniques for solving surface integral examples?

Some common techniques for solving surface integral examples include parameterization, where the surface is represented by a set of equations in terms of two parameters, and using vector calculus to simplify the integrand. Other techniques include using symmetry to reduce the complexity of the integral and changing variables to transform the integral into a simpler form.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
546
  • Calculus and Beyond Homework Help
Replies
6
Views
988
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
Back
Top