Triple Integral in Spherical Coordinates

AI Thread Summary
For a hemispherical surface of radius R, the polar angle theta must be restricted to 0 < theta < pi/2 to ensure only points above the x-y plane are included. The radius ρ ranges from 0 to R, while the azimuthal angle phi spans from 0 to 2π. To calculate the triple integral of the divergence of a vector field v over this volume, the formula used is ∭v·dV = ∭(ρ^2sinθ)(∇·v) dρdθdφ. Substituting the divergence of v into this formula allows for evaluation with the specified bounds. This approach effectively addresses the integration over the defined hemispherical volume.
cepheid
Staff Emeritus
Science Advisor
Gold Member
Messages
5,197
Reaction score
38
I have a hemispherical surface of radius R with it's base centred on the origin. We are using the convention:

r is the radius i.e. the magnitude of the position vector of a point: its distance from the origin.

theta is the polar angle

phi is the azimuthal angle

I am asked to calculate the integral of the divergence of a given vector field v over this volume (enclosed by the hemisphere).

So far what I have done is to say that a full sphere would be given by the equation:

r = R

What about a hemisphere? It seems to me that the angle theta must be restricted so that points below the x-y plane are not part of the domain. So what are the allowed values of theta? I can't seem to figure out whether it should be -pi/2 < theta < pi/2, or something else? I need to know this to set my bounds of integration for one of the three integrals. Thanks.
 
Last edited:
Physics news on Phys.org
For a hemisphere : r goes from 0 to R
theta goes from : 0 to 90°
phi goes from : 0 to 360°

Ofcourse, you know that you need to express the angles in radials

regards
marlon
 


The allowed values of theta for a hemisphere are 0 < theta < pi/2. This is because the polar angle theta measures the angle from the positive z-axis, and a hemisphere only includes points above the x-y plane. Therefore, the lower bound of theta should be 0, and the upper bound should be pi/2.

To calculate the triple integral in spherical coordinates, we can use the formula:

∭v·dV = ∭(ρ^2sinθ)(∇·v) dρdθdφ

Where ρ is the distance from the origin, θ is the polar angle, and φ is the azimuthal angle.

In this case, since the vector field v is given, we can calculate ∇·v and substitute it into the formula. Then, we can use the bounds of integration mentioned above to evaluate the triple integral over the hemisphere.

It is also important to note that the radius ρ should range from 0 to R, as we are only considering points within the hemisphere of radius R.

Overall, the triple integral in spherical coordinates for this scenario would be:

∭v·dV = ∭(ρ^2sinθ)(∇·v) dρdθdφ

where the bounds of integration are: 0 ≤ ρ ≤ R, 0 ≤ θ ≤ π/2, 0 ≤ φ ≤ 2π.

I hope this helps with your calculations. Good luck!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top