Mass of a sphere with a vertical bound

In summary, to find the total mass occupying a solid region D bounded by a sphere of radius 3 centered at the origin and z = 1, with a density function of 1/1+x^2+y^2+z^2, you can use Cartesian coordinates for easier calculation. The bounds for x, y, and z would be -sqrt(9-y^2-z^2) to sqrt(9-y^2-z^2), -sqrt(9-z^2) to sqrt(9-z^2), and -3 to 1, respectively. The integral for the first coordinate would be 1/(1+x^2+y^2+z^2)dx evaluated from -sqrt(9-y^2-z
  • #1
Harlow
2
0
Find the total mass that occupies a solid region D bounded by a sphere of radius 3
centered at the origin and z = 1 if the density of the function is (x, y, z) = 1/1+x^2+y^2+z^2 .








I would like to be able to do this problem using spherical coordinates but I am unsure about how this z=1 will affect my bounds of integration for both phi and rho.
 
Physics news on Phys.org
  • #2
Welcome to PF!

Hi Harlow! Welcome to PF! :smile:

(have a theta: θ and a phi: φ and a rho: ρ :wink:)

Personally, I'd use Cartesian coordinates or cylindrical coordinates, but if you must use spherical coordinate, draw a radius at angle θ to the z-axis, and see where the limits of r are along that radius. :smile:
 
  • #3
Thanks for the reply. To be honest I hadn't even really considered using Cartesian coordinates. We have been covering spherical and cylindrical coordinates so I figured that the problem was probably designed for one of those two coordinate systems. Cartesian looks easier for this problem. Heres what I have for the bounds x: -sqrt(9-y^2-z^2) to sqrt(9-y^2-z^2), y: -sqrt(9-z^2) to sqrt(9-z^2) and z:-3 to 1.

My only issue is with the integral, I have the first integral set up as follows: 1/(1+x^2+y^2+z^2)dx evaluated from -sqrt(9-y^2-z^2) and sqrt(9-y^2-z^2). does this look correct? How can I do this integral?
 
Last edited:

FAQ: Mass of a sphere with a vertical bound

What is the formula for calculating the mass of a sphere with a vertical bound?

The formula for calculating the mass of a sphere with a vertical bound is: mass = density x volume. This formula applies to all types of spheres, including those with a vertical bound.

How do you determine the density of a sphere with a vertical bound?

The density of a sphere with a vertical bound can be determined by dividing its mass by its volume. The mass can be measured using a scale, and the volume can be calculated using the formula for the volume of a sphere: V = (4/3)πr³, where r is the radius of the sphere.

Can you explain the concept of a vertical bound in relation to a sphere?

A vertical bound refers to a restriction or limitation on the vertical movement of a sphere. This could be a physical barrier or a mathematical boundary, such as a specific height or depth. It is important to consider the vertical bound when calculating the mass of a sphere, as it can affect the volume and density of the sphere.

Is there a difference in the mass of a sphere with a horizontal bound versus a vertical bound?

Yes, there can be a difference in the mass of a sphere with a horizontal bound versus a vertical bound. This is because the vertical bound can affect the volume and density of the sphere, which in turn affects the mass. For example, a sphere with a vertical bound may have a smaller volume and higher density than a sphere with a horizontal bound, resulting in a different mass.

How does the mass of a sphere with a vertical bound affect its gravitational pull?

The mass of a sphere with a vertical bound does not directly affect its gravitational pull. Gravitational pull is determined by the mass and distance of an object. However, the mass of a sphere with a vertical bound can indirectly affect its gravitational pull if it is in the vicinity of other objects with a gravitational pull, as the mass of the sphere can contribute to the overall gravitational force.

Back
Top