Would the cone angle be 1 radian (180/pi degrees) ?

Not answered, but nice illustration here: http://www.squ1.com/light/quantities.html

A unit of measure equal to the solid angle subtended at the center of a sphere by an area on the surface of the sphere that is equal to the radius squared: The total solid angle of a sphere is 4π (4 pi)steradians.

The unit of solid angle adopted under the System International d'Unites.

I don't know if a surface of revolution will help in this problem as it calculates a volume.  Anyway it works like this.

     Make a function that models the outer edge of an object that is three dimentional for a cartesian coordinate system.  After you have the function it is simply a matter of integration.

     If y=f(x) the Volume would be ∫2pixf(x) dx] evaluated from 0 to the radius.  (for a solid object)  Perhaps an example is in order.

     Take a cone.  If you look at the right half of it the side will match a decreasing line with the y intercept being the apex.

     The equation for that line will be y = H - (H/R)x, this is true for all cones.  This line intercepts the x axis at x = R.

     Now multiply your line function by (2 pi x) and then integrate from zero to R.  You will get:

     2pix(H - (H/R)x) = 2piH ∫x - x^2 / R dx  (remember to factor out constants?)

     2piH [ x^2 / 2 - x^3 / 3R ] from 0 to R

     2piH (R^2 / 2 - R^3 / 3R)

     2piH ( 3R^2 - 2R^2 ) / 6

     This all equals piHR^2 / 3

     This is the common formula for the area of a cone.  This is the same way that the volume of a sphere was calculated using the curve of one quadrant and multiplying the integration by two to add the bottom half.  You can use this method for any object with a little imagination. 

     Note that we used y = f(x) because we were rotating around the y axis.  If you rotate around the x axis you will need a function like x = f(y).

     You can also calculate surface area if at the end of your computation you divide by the thickness or radius of the object in question.  If you want a volume of an object that is not solid you simply subtract the integration of the inner boundary from that of the outer boundary.

     I hope this helps and would love if you post any uncommon volumes that you figure out in the future.

edit:  I am sure that you understand that an integration can be interpreted as the area below the curve or line.  What we have done above is that we have integrated that area for every value of the radius also.  That is why we multiply the function by 2pix (circumference formula) or 360 degrees.

Last edited by irspow (2005-12-10 09:33:26)

(Put a ":" in front of "pi" and you get π, see http://www.mathsisfun.com/forum/viewtop … 623#p17623)

John, I was trying to verify my suggestion that the interior angle was indeed correct at 90/π²°.  This was because if you multiply this by 360° you will get the amount of square degrees in a steradian.

  As it turns out I am having a very difficult time integrating the formula to prove it.  I have to integrate ∫2πf(y)√1+(dx/dy)²  dy.

  I tried integration by parts and trigonomic substitution to no avail.

  The integration is the general form for a surface of revolution about the y axis.  I think that that is what you need for this proof.

  f(y) = √1-y²  is what needs to be applied to the formula above.  Without going much further at this point the upper and lower bounds are no problem here but do depend on how the integration is solved.

  So if anyone can solve the integration above for the general case, we could all finally put an end to the mystery of how exactly square degrees are related to steradians within three dimensional space.  Because once the integration above is completed and evaluated over the correct range it should simply equal 1.

Yes John, that is precisely what I was trying to write.  As far as my efforts to figure this thing out this is what I have come up with.

   The problem that I was having with the integration was silly after realizing that I was making substitutions too early complicating the matter more than was necessary.

   f(y) = √1-y²

   When I first saw this I immediately start using the trigonomic substitution which made this completely unworkable, at least for me.

   Leaving f(y) alone until I had the integrand simplified as much as possible was the solution for me.  Sadly it took a lot of crumpled pieces of paper and a lot of time pouring over the integration tables to figure this out.

   I solved it easily after that point.  Just setting u = y². (smacking myself!)  Then I was reproached as the surface area for the angle that I proposed earlier was only .019883871 units²!

   So then I solved the integral backwards to find the true angle within the sphere at the apex.  Basically I just solved for when the integral equalled one. (That was what was stated in your original post)

   This solution is only half of the interior angle because of the way the surface formula works.  The angle that solved the integral was about 32.77°.  So the whole angle would be about 65.541°.  It turns out my theory of 360° multiplied by the interior angle was further off than your original guess from the beginning.

   I could not find a relationship between this interior angle and the square degrees, and I tried very hard to believe me.  That 65.541° is reliable though because that integral is used to calculate just the type of thing that we are doing here.

   By the way you should note that if you remove the 2πf(y) from the integrand it will now solve for the length of the arc!  You can use either f(x) or f(y) depending on which is more convenient for the length of arc.

   I hope all of this jabbering that I've done on this thread has helped you somewhat, I know that I have learned a thing or two since I first ventured into this topic.

   Oh!  Is there somewhere that I can learn about that [math] syntax?

Last edited by irspow (2005-12-12 11:25:29)

I saw that you posted in another thread about the arc length integral, very good. (useful too) Try to remember that one it comes in handy ofter.  I actually solved the integral that I was having so much trouble with.

   As funny as it sounds, in the end it was nothing more than 2π (√u) evaluated from cos²0° to cos²32.77053659° which provided the answer of 1 square unit.  The u came from a substitution for y² after much simplification.

   Anyway, I still haven't found a relationship between this interior angle (65.54107318° which is twice the half angle used for the surface of revolution) and the 3282.80635°² in one steradian.  I think it is because there is no relationship to be found.  By definition a 4π steradians = 4πr².

   Sorry that is the best that I can do on this one for now.

Last edited by irspow (2005-12-12 15:21:05)