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Hyperbola and Cone


Date: 03/04/99 at 10:04:28
From: Mikael
Subject: Hyperbola and Cone

I have some problems in mathematics.

1) How do you calculate the maximum and minimum distance from (0,0,0) 
   to the surface x^2+y^2+z^2+x*y = 1 and the volume that the 
   expression spans?

2) A cone has radius 2 dm and height 2 dm. A hole with radius 1 dm and 
   length 2 dm is made in the cone, so the edge of the hole tangents 
   the symmetric axis of the cone. Calculate the volume and the area of 
   the hole.

Thnanks for your help in advance!


Date: 03/04/99 at 15:43:50
From: Doctor Rob
Subject: Re: Hyperbola and Cone

In the first problem, you can rotate the surface about the z-axis by
Pi/4 by making the substitution

   x = (X+Y)/sqrt(2)
   y = (X-Y)/sqrt(2)
   z = Z

and get the equation

   X^2/(2/3) + Y^2/2 + Z^2 = 1

This is the equation of an ellipsoid centered at the origin, with
semi-axes sqrt(2/3), sqrt(2), and 1.  The max and min distances are now 
easy, and the volume enclosed by it is 4*Pi/3 times the product of 
these three.

I do not think I understand the second question. I guess you want the
volume and surface area of the intersection of the interiors of the
cone and a cylinder so described. The only way I know to do this is
to use calculus. The equation of the cone is

   x^2 + y^2 - (z-2)^2 = 0,

and the equation of the cylinder is

   (x-1)^2 + y^2 = 1.

(The axis of the cone is x = y = 0, the z-axis, which lies within the 
surface of the cylinder.) Then the volume of the intersection is most 
easily found using cylindrical coordinates. Set

   x = r*cos(t)
   y = r*sin(t)

Then the cone, its base, and the cylinder have equations

   z = 2 - r
   z = 0
   r = 2*cos(t)

Then the enclosed region is described by the following inequalities:

   -Pi/2 <= t <= Pi/2
   0 <= r <= 2*cos(t)
   0 <= z <= 2 - r

Now to get the volume, we integrate

        Pi/2  2*cos(t)  2-r
   V =  INT     INT     INT r dz dr dt
       -Pi/2     0       0

        Pi/2  2*cos(t)
     =  INT     INT    2*r-r^2 dr dt
       -Pi/2     0

        Pi/2
     =  INT  4*cos^2(t) - (8/3)*cos^3(t) dt
       -Pi/2

which you can finish. To get the surface area, we see that on the
surface, r = 2*cos(t), so

   -Pi/2 <= t <= Pi/2
   x = 2*cos^2(t)
   y = 2*cos(t)*sin(t)
   0 <= z <= 2 - 2*cos(t)

Then,

   dx/dt = -4*sin(t)*cos(t) = -2*sin(2*t)
   dy/dt = 2*cos^2(t)-2*sin^2(t) = 2*cos(2*t)
   sqrt[(dx/dt)^2 + (dy/dt)^2] = 2

The we integrate

        Pi/2  2-2*cos(t)
   S =  INT      INT     sqrt[(dx/dt)^2 + (dy/dt)^2] dz dt
       -Pi/2      0

        Pi/2  2-2*cos(t)
   S =  INT      INT     2 dz dt 
       -Pi/2      0

        Pi/2
     =  INT   4 - 4*cos(t) dt
       -Pi/2

which you can finish.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
High School Calculus

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