Mass of a Sphere: Solve w/ Triple Integrals & Spherical Coordinates

In summary, the question asks to find the mass M of a sphere of radius a, with a mass density that is proportional to the distance from the center of the sphere. The solution involves using triple integrals in spherical coordinates, with the variable r representing the distance from the center. The mass can be calculated by integrating Kρ^3sin(θ) with respect to r, where K is a constant.
  • #1
reminiscent
131
2

Homework Statement


Find the mass M of a sphere of radius a, if its mass density is proportional to the distance
from the center of the sphere.

Homework Equations


Triple integrals using spherical coordinates

The Attempt at a Solution


The only place where I am stuck is if the density is KpcosΦ or just Kp. So is it integrating KpcosΦp2sinΦ or Kp3sinΦ?
 
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  • #2
reminiscent said:

Homework Statement


Find the mass M of a sphere of radius a, if its mass density is proportional to the distance
from the center of the sphere.

Homework Equations


Triple integrals using spherical coordinates

The Attempt at a Solution


The only place where I am stuck is if the density is KpcosΦ or just Kp. So is it integrating KpcosΦp2sinΦ or Kp3sinΦ?
Let me emphasize a part of the question: its mass density is proportional to the distance from the center of the sphere.
 
  • #3
Hi reminiscent:

You only need to integrate with respect to r. What is the mass of a shell of thickness dr at radius r?

Regards,
Buzz
 
  • #4
can you guys please elaborate on this?
 
  • #5
Samy_A's point is that the problem said that the mass is proportional to the distance to the center of the sphere. That distance is the variable [itex]\rho[/itex].

BuzzBloom's point is that, since the mass is given by [tex]\int_0^a\int_0^\pi\int_0^{2\pi} K\rho (\rho^2 sin(\theta) d\phi d\theta d\rho)= \int_0^a\int_0^\pi\int_0^{2\pi} K\rho^3 sin(\theta) d\phi d\theta d\rho[/tex]
 
  • Like
Likes Samy_A
  • #6
thanks :D
 

1. What is the formula for calculating the mass of a sphere using triple integrals and spherical coordinates?

The formula for calculating the mass of a sphere using triple integrals and spherical coordinates is given by:
M = ρ ∫∫∫ r2sinφ dr dφ dθ
where ρ is the density function, r is the distance from the origin, φ is the angle between the positive z-axis and the position vector, and θ is the angle between the positive x-axis and the projection of the position vector onto the xy-plane.

2. What are the advantages of using triple integrals and spherical coordinates to find the mass of a sphere?

Using triple integrals and spherical coordinates allows for a more efficient and accurate calculation of the mass of a sphere compared to other methods. This is because spherical coordinates are better suited for working with spherical shapes and the triple integral allows for the integration of multiple variables at once, reducing the number of steps required in the calculation.

3. Can you walk me through the steps of solving for the mass of a sphere using triple integrals and spherical coordinates?

Sure, the first step is to set up the triple integral using the formula:
M = ρ ∫∫∫ r2sinφ dr dφ dθ
Next, determine the limits of integration for each variable based on the bounds of the sphere. Then, evaluate the integral using appropriate techniques such as substitution or integration by parts. Finally, multiply the result by the density function to obtain the mass of the sphere.

4. Can this method be applied to calculate the mass of any type of sphere?

Yes, this method can be applied to calculate the mass of any type of sphere, as long as the density function is known and the sphere can be represented in spherical coordinates.

5. How accurate is the result obtained from using triple integrals and spherical coordinates to find the mass of a sphere?

The result obtained from using triple integrals and spherical coordinates to find the mass of a sphere is highly accurate, as long as the limits of integration and the density function are correctly determined. Any errors in the calculation would most likely be due to human error rather than the method itself.

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