Triple Integral in Spherical Coordinates

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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.
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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.
 
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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!
 
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