Expressing a surface in cartesian coordinates from spherical

Click For Summary
To convert the surface defined by θ = π/4 from spherical to Cartesian coordinates, the relationship between the coordinates must be established using the equations x = r sin(θ) cos(Ø), y = r sin(θ) sin(Ø), and z = r cos(θ). Given θ = π/4, the equation simplifies to z = r cos(π/4), which leads to z = (1/√2)r. The value of r can vary from 0 to infinity, while Ø can range from 0 to 2π. It's crucial to ensure that the notation used for the spherical coordinates is consistent, as any discrepancies can affect the calculations. The final expression represents a plane in three-dimensional space where z is proportional to the radial distance r.
acedeno
Messages
34
Reaction score
4

Homework Statement


The following equation describes a surface in spherical coordinates. θ =pi/4
Write the equation in the cartesian coordinates?

that is, (r,θ,Ø) to (x,y,z)

Homework Equations


x=rsinθcosØ
y=rsinθsinØ
z=rcosθ

r=sqrt(x^2+y^2+z^2)
θ=cos^-1(z/r)
Ø=tan^-1(y/x)


The Attempt at a Solution


I'm pretty stumped. The only start I can get is that if θ=pi/4, this means that r and phi have to work around this. Allowing r to be from 0 to infinity. I'm not really sure what happens to phi. I'm thinking that it can be anywhere from 0 to 2pi.
 
Physics news on Phys.org
If your surface is simply defined as

\theta = \pi/4

then all you need to do is solve your conversion factor from theta of

\theta = cos^{-1}(\frac{z}{r}) = cos^{-1}(\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}})

Since your value of theta is a constant, you just have

\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}} = cos(\frac{\pi}{4})

Although you may want to make sure you have your notation correct. Generally I've seen

z = r cos\phi, x = r cos\theta sin\phi, y = r sin\theta sin\phi

and if you did in fact get your notation mixed up (which it's possible that you did or didn't, but it's worth checking) that changes the nature of your calculations.
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
866
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
943