Question

Surfaces of revolution

Answer 1

Hey so I have a question, my prof says cross sections don't always give you the right answers, and he does these quadratic surfaces a weird way, which seems arbitrary to me, so for example what if we have a problem like this \[x^2-y^2-z^2=1\] I would have assumed you would make each variable a constant and trace from there but it's not right I guess...
@ganeshie8 @Kainui

Answer 2

So the way I learned it, I would first \[x^2-(y^2+z^2)=1\] since we know \[x^2-y^2=1\] is a hyperbola. Then we would rotate about the x - axis, but I think there is some polar coordinates implied somewhere here.

Answer 3

Why exactly are we choosing the x - axis to rotate it around, or is that arbitrary to.

Answer 4

u can say y^2+z^2 = r^2

Answer 5

Right!

Answer 6

and u have a hyperbola wrt to the radius so a hyperbola at every angle

Answer 7

so same as rotating one hyperbola for some cross sectional plane or just take
z=0
then it will be x^2-/y^2
in this case r^2 just happened to be y^2
this be the same it will be rorated at every angle

Answer 8

That's it haha? So is that what we do with most of the questions related to as such.

Answer 9

i dont think so its my first time seeing this trick im sure theres other ways

Answer 10

do u have another qusetion like this?

Answer 11

Mhm, I'm trying to look for one

Answer 12

Ok what if we had something like \[3x^2-2y^2+4z^2=1\]

Answer 13

elliptical hyperbola?

Answer 14

u can stick
(3x^2+4z^2) -2y^2 =1

Answer 15

Yeah, that's what I'm thinking

Answer 16

then do a change of coordinates to rescale the ellipse into a circle and u get the same scenario

Answer 17

and use the jacboian to conintue to work with integration and stuff for all useful purposes that is enuff but if ur prof wants its to be drawn only on x y z, and the picture is the most important then

Answer 18

Ok well lets make it easier on ourselves just for a moment so I understand this process haha, lets make the coefficients the same for x^2+z^2

Answer 19

ya then tis the same thing but now

Answer 20

u rotate about the R axis?

Answer 21

lemme think

Answer 22

Ok so (x^2+z^2)-2y^2 = 1 then here I'd be rotating it around the y-axis right

Answer 23

is it y
for
x^2-r^2 =1 u rottated about the x axis

Answer 24

for r^2-y^2=1?

Answer 25

Yeah pretty much

Answer 26

No for z^2-y^2=1 is rotated around x axis if that's what you mean

Answer 27

im thinking its a hyperbola but now its rotated polarly

Answer 28

it is a hyperbola

Answer 29

so u get like a hour glass

Answer 30

hmm

Answer 31

Yeah basically, I'm thinking something like this