Triple Integral in Spherical Coordinates

In summary, the conversation is about calculating the integral of the divergence of a given vector field over a volume enclosed by a hemisphere with radius R and centered on the origin. The convention for the variables r, theta, and phi are explained, and it is mentioned that for a hemisphere, r goes from 0 to R, theta goes from 0 to 90 degrees, and phi goes from 0 to 360 degrees. The importance of expressing the angles in radians is also emphasized.
  • #1
cepheid
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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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  • #2
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
 
  • #3


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!
 

1. What is a triple integral in spherical coordinates?

A triple integral in spherical coordinates is a mathematical tool used to calculate the volume of a three-dimensional object. It involves integrating over a region in spherical coordinates, which are represented by the three variables: radius (r), inclination angle (θ), and azimuthal angle (φ).

2. How is a triple integral in spherical coordinates different from other coordinate systems?

A triple integral in spherical coordinates differs from other coordinate systems, such as cartesian or cylindrical coordinates, in that it uses spherical coordinates to represent points in three-dimensional space. This allows for a more intuitive approach to calculating volume, as it takes into account the shape of the object being integrated over.

3. What are the advantages of using a triple integral in spherical coordinates?

One advantage of using a triple integral in spherical coordinates is that it simplifies the calculation of volumes for objects with spherical symmetry. Additionally, it can be easier to visualize and set up the integral in spherical coordinates for certain types of objects, such as spheres or cones.

4. What are the limitations of using a triple integral in spherical coordinates?

One limitation of using a triple integral in spherical coordinates is that it may not be the most efficient method for calculating volume for objects that do not have spherical symmetry. Additionally, it may be more difficult to set up the integral in spherical coordinates for certain types of objects, such as cylinders or cubes.

5. How can a triple integral in spherical coordinates be applied in real-world scenarios?

A triple integral in spherical coordinates can be applied in real-world scenarios to calculate the volume of three-dimensional objects, such as spherical tanks or domes. It can also be used in physics and engineering to calculate the moments of inertia for rotating objects or the electric fields of charged spherical objects.

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