MHB How Do You Calculate the Normal Vector of a Sphere in Spherical Coordinates?

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To calculate the normal vector of a sphere defined by V = x^2 + y^2 + z^2 ≤ 1 in spherical coordinates, the conversion to spherical coordinates is given as x = r sin(t) cos(f), y = r sin(t) sin(f), and z = r cos(t), with t ranging from 0 to π and f from 0 to 2π. The normal vector can be found by taking the gradient of the surface function f(x, y, z) = 0. For this specific case, the discussion also mentions the intention to prove the divergence theorem of Gauss. It is noted that the method remains the same when using polar coordinates, requiring the gradient in those coordinates instead of Cartesian ones. Understanding these concepts is essential for further calculations and proofs related to the sphere.
brunette15
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I am given the sphere V= x^2 + y^2 + z^2 =< 1

I have converted it to spherical coordinates:

x = rsin(t)cos(f)
y = rsin(t)cos(f)
z = rcos(t)

where t ranges from 0 to pi, and f ranges from 0 to 2pi.

I am unsure how to go about this problem from here. Any guidance would be really appreciated :)
 
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brunette15 said:
I am given the sphere V= x^2 + y^2 + z^2 =< 1

I have converted it to spherical coordinates:

x = rsin(t)cos(f)
y = rsin(t)cos(f)
z = rcos(t)

where t ranges from 0 to pi, and f ranges from 0 to 2pi.

I am unsure how to go about this problem from here. Any guidance would be really appreciated :)

Hey brunette15! ;)

We can find the normal of a surface $f(x,y,z)=0$ by taking the gradient $\nabla f$... (Thinking)
 
I like Serena said:
Hey brunette15! ;)

We can find the normal of a surface $f(x,y,z)=0$ by taking the gradient $\nabla f$... (Thinking)

Hi again I like Serena :)

I am aware that we could do that but for this particular case i am trying to prove the divergence theorem of Gauss :/
 
brunette15 said:
Hi again I like Serena :)

I am aware that we could do that but for this particular case i am trying to prove the divergence theorem of Gauss :/

Huh? :confused:

What do you want to do then?

Do you want to find the normal in polar coordinates?
If so, the method is the same - we just need the gradient in polar coordinates instead of cartesian coordinates.
 
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