Solving Vector Area dS of Sphere: Find Mistake!

In summary, the conversation is about someone trying to calculate the vector area of a sphere but getting mixed up with their results. They ask for help in finding their mistake and provide their attempted solution, which includes a parametrization of a sphere and a calculation using partial derivatives. After realizing their mistake, they correct the parametric equation and conclude the conversation.
  • #1
aaaa202
1,169
2

Homework Statement


I wonna calculate the vector area dS of a sphere, but for some reason my result gets mixed areound. I need a trained eye to see where I make a silly mistake.


Homework Equations


Parametrization of a sphere:

(x,y,z) = r(cos[itex]\phi[/itex]sin[itex]\theta[/itex],sin[itex]\phi[/itex]cos[itex]\theta[/itex],cosθ)

The Attempt at a Solution


So ∂r/∂[itex]\phi[/itex] = r(cosθcos[itex]\phi[/itex],-sinθsin[itex]\phi[/itex],sinθ)
and
∂r/∂θ = r(-sin[itex]\phi[/itex]sinθ,cos[itex]\phi[/itex]sinθ,0)

and dS = ∂r/∂[itex]\phi[/itex] x ∂r/∂θ = r2 (sinθcosθcos[itex]\phi[/itex],sin[itex]\phi[/itex]sinθ2, ...) dθd[itex]\phi[/itex]

But the first two terms should be switched around according to my notes! Where do I go wrong? :(
 
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  • #2
aaaa202 said:

Homework Statement


I wonna calculate the vector area dS of a sphere, but for some reason my result gets mixed areound. I need a trained eye to see where I make a silly mistake.


Homework Equations


Parametrization of a sphere:

(x,y,z) = r(cos[itex]\phi[/itex]sin[itex]\theta[/itex],sin[itex]\phi[/itex]cos[itex]\theta[/itex],cosθ)

The Attempt at a Solution


So ∂r/∂[itex]\phi[/itex] = r(cosθcos[itex]\phi[/itex],-sinθsin[itex]\phi[/itex],sinθ)
and
∂r/∂θ = r(-sin[itex]\phi[/itex]sinθ,cos[itex]\phi[/itex]sinθ,0)

and dS = ∂r/∂[itex]\phi[/itex] x ∂r/∂θ = r2 (sinθcosθcos[itex]\phi[/itex],sin[itex]\phi[/itex]sinθ2, ...) dθd[itex]\phi[/itex]

But the first two terms should be switched around according to my notes! Where do I go wrong? :(

Well, you have the ##\theta## and ##\phi## reversed from the usual math notation for spherical coordinates, but I guess some physicists do that. But when you differentiated with respect to ##\phi## it looks like you differentiated the ##\theta## variables, and conversely.
 
  • #3
nvm.. Found my mistake - the parametric equation was wrong.
 

1. What is the formula for finding the vector area dS of a sphere?

The formula for finding the vector area dS of a sphere is dS = r^2sinθdθdφ, where r is the radius of the sphere, θ is the angle between the vector and the z-axis, and φ is the angle between the vector and the x-axis.

2. What is the mistake in the given problem?

The mistake in the given problem is that the formula used to find the vector area dS is incorrect. It is missing the sinθ term, which is crucial in calculating the correct vector area of a sphere.

3. How can the mistake be corrected?

The mistake can be corrected by adding the sinθ term to the formula, as follows: dS = r^2sinθdθdφ. This will give the correct vector area of a sphere.

4. What is the importance of finding the correct vector area of a sphere?

Finding the correct vector area of a sphere is important in many applications of physics and mathematics. It is used to calculate the flux of a vector field through a sphere, which is essential in understanding the behavior of electric and magnetic fields, fluid dynamics, and many other physical phenomena.

5. Can the formula for finding the vector area dS of a sphere be applied to other shapes?

Yes, the formula dS = r^2sinθdθdφ can be applied to other shapes, such as cones, cylinders, and ellipsoids, as long as the appropriate values for r, θ, and φ are used. However, it is important to note that for other shapes, the formula may vary and should be checked for accuracy.

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