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/
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