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

[fayled]

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.

 

[Dick]

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.

 

[fayled]

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?

 

[The Dog Star]

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.

 

[HallsofIvy]

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.

 

[fayled]

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!

 

[LCKurtz]

There's a picture of a similar problem at
https://www.physicsforums.com/showthread.php?t=734672#post4640511