Transforming from cartesian to cylindrical and spherical

In summary: The right side of your spherical equation could be simplified by removing the terms involving φ and θ. This would leave you with: ρcosφ + ρsinφcosθ = 2ρsinφ and ρcosφ + ρsinφsinθ = 2ρcosφ.
  • #1
yango_17
60
1

Homework Statement


Translate the following equations from the given coordinate system into equations in each of the other two systems. Also, identify the surfaces so described by providing appropriate sketches.

Homework Equations

The Attempt at a Solution


For my solutions, I obtained z=2r^2 for the cylindrical equation and for the spherical equation I got:
(ρcosφ)^2 = 2(ρsinφcosθ)^2 + 2(ρsinφsinθ)^2. For my sketch I drew an infinite cylinder:
DoubleCone.png

I was wondering whether my conversions were correct, as when I transform the same equation from cartesian to spherical and from cylindrical to spherical I seem to obtain different equations.
 
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  • #2
yango_17 said:

Homework Statement


Translate the following equations from the given coordinate system into equations in each of the other two systems. Also, identify the surfaces so described by providing appropriate sketches.

Homework Equations

The Attempt at a Solution


For my solutions, I obtained z=2r^2 for the cylindrical equation and for the spherical equation I got:
(ρcosφ)^2 = 2(ρsinφcosθ)^2 + 2(ρsinφsinθ)^2. For my sketch I drew an infinite cylinder:
DoubleCone.png

I was wondering whether my conversions were correct, as when I transform the same equation from cartesian to spherical and from cylindrical to spherical I seem to obtain different equations.
That's not a cylinder -- it's a cone. (Mathematically, cones have two parts.)

What was your equation in Cartesian form?
 
  • #3
Sorry I meant cone hahah. My original equation in cartesian form was z^2=2x^2+2y^2
 
  • #4
yango_17 said:
Sorry I meant cone hahah. My original equation in cartesian form was z^2=2x^2+2y^2
Then your cylindrical equation should be ##z^2 = 2r^2##, not ##z = 2r^2## as you showed earlier.
 
  • #5
Does the spherical equation look correct?
 
  • #6
I'll take a look at it in a little while (it's dinner time...)
 
  • #7
Much appreciated (:
 
  • #8
yango_17 said:
(ρcosφ)^2 = 2(ρsinφcosθ)^2 + 2(ρsinφsinθ)^2
This is correct, as far as you went, but the right side could be simplified considerably.
 
  • #9
How would you go about simplifying it?
 
  • #10
yango_17 said:
How would you go about simplifying it?
Expand the terms on the right side.
 

What is the difference between cartesian, cylindrical, and spherical coordinates?

Cartesian coordinates use three perpendicular axes (x, y, z) to describe the position of a point in space. Cylindrical coordinates use two perpendicular axes (r, z) and an angle (θ) to describe a point in space. Spherical coordinates use a radial distance (r), an angle (θ), and an azimuth angle (φ) to describe a point in space.

Why would one need to transform from cartesian to cylindrical or spherical coordinates?

Transforming from cartesian to cylindrical or spherical coordinates can make certain calculations or visualizations easier, depending on the problem at hand. For example, cylindrical coordinates are useful for solving problems with circular symmetry, while spherical coordinates are useful for solving problems with spherical symmetry.

What is the formula for converting from cartesian to cylindrical coordinates?

The formula for converting from cartesian to cylindrical coordinates is:

r = √(x2 + y2)
θ = tan-1(y/x)
z = z

Is there a difference in the formulas for converting to cylindrical coordinates versus spherical coordinates?

Yes, the formulas for converting to spherical coordinates are slightly different:

r = √(x2 + y2 + z2)
θ = tan-1(y/x)
φ = cos-1(z/√(x2 + y2 + z2))

Can I transform from cylindrical or spherical coordinates back to cartesian coordinates?

Yes, you can use the same formulas for converting from cartesian to cylindrical or spherical coordinates, but in reverse. Here are the formulas for converting back to cartesian coordinates:

x = r cos(θ)
y = r sin(θ)
z = z

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