Question

how to write x^2=16-z^2 in cylindrical and spherical coordinates??

Answer 1

for cylindrical coordinates.
z = z
x = rcos\theta
y = rsin\theta

http://en.wikipedia.org/wiki/Cylindrical_coordinate_system#Cartesian_coordinates
put those values in place of x,y,z then you will have it.

Answer 2

for spherical coordinates,
z = rcos\theta
x = r sin\theta * cos\phi
y = r sin\theta * sin\phi

Answer 3

means we can't find the value of r right?just replace x=r cos theta into the equation...and the new equation contains r,theta and z...am i right?

Answer 4

yes we can find the value of r, r = sqrt(x^2+y^2), and theta = arctan(y/x)

but these are going to be our new variables instead of x,y, z

Answer 5

what type of surface we get from equation (y^2/25)-(x^2/4)=z/3??

Answer 6

loks like hyperboloid

Answer 7

it was called hyperbolic paraboloid
https://www.google.com/search?q=hyperbolic+paraboloid&hl=en&rlz=1C1ASUT_enNP447NP447&prmd=imvns&tbm=isch&tbo=u&source=univ&sa=X&ei=Vz-QT9O1D8nqrQfM49npBA&ved=0CEkQsAQ

Answer 8

how you know the surface is hyperbolic paraboloid based on the equation?

Answer 9

The surface is a cylinder around the y-axis with radius 4

Answer 10

google + wolfram
plus (y^2/25)-(x^2/4) looks like hyperbola.

Answer 11

The quation is
\[
x^2 + z^2 = 4^2
\]
Any plan perpendiula to the y-axis cus th surfacei ircle of radus 4

Answer 12

Any plan perpendiular to the y-axis cus thr surface in a circle of radus 4

Answer 13

yes it is, but how do we do for (y^2/25)-(x^2/4) = z/3

Answer 14

But, we are not talking about this equation now. We are talking about
\[ x^2 + z^2=16
\]

Answer 15

yeah ... I understand that.

Answer 16

(lol)