Associated Topics || Dr. Math Home || Search Dr. Math

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.

```

```
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

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search