Find the center of mass of the hemisphere x^2+y^2+z^2=a^2, z ⩾0, if it has constant density.

Name: Problem 37, Chapter 13: Essential Calculus Early Transcendentals
Uploaded: 2020-07-04T21:20:25-08:00
Duration: 6 min 7 s
Channel: Wesley Hines
Description: Find the center of mass of the hemisphere x^2+y^2+z^2=a^2, z ⩾0, if it has constant density.

Find the center of mass of the hemisphere x^2+y^2+z^2=a^2, z...

Key Concepts

Constant Density

When an object has constant density, the density value can be factored out of the integrals for both mass and moment calculations. This simplification reduces the problem to pure geometric integration, where the shape's structure and boundaries dictate the resulting mass distribution.

Volume Integration

Volume integration is a key calculus method used for summing infinitesimal mass elements over the whole object. For continuous mass distributions, the mass and moments are calculated by integrating the density multiplied by the differential volume element, allowing one to determine the overall mass and the coordinates of the center of mass.

Spherical Coordinates

Spherical coordinates are a coordinate system particularly well-suited to problems involving spheres or hemispheres. By expressing the position in terms of radial distance, polar angle, and azimuthal angle, the boundaries of integration align naturally with the geometry of spherical objects, and the differential volume element takes a specific form that simplifies the integration.

Center of Mass

The center of mass is the point in a body or system of particles where the total mass of the system may be considered to be concentrated. Mathematically, it is calculated by taking a weighted average of the position vectors of all mass elements, which involves integrating the product of the position and the density over the entire volume and then dividing by the total mass.

Symmetry

Symmetry in a mass distribution can significantly simplify the computation of the center of mass. If the object exhibits symmetry with respect to a plane or axis, the center of mass must lie along that axis or in the plane. In cases with uniform density, these symmetric properties help eliminate components that would otherwise require integration.

Transcript

00:01 So the center of mass, which we write as x bar, y, bar, z bar, is given by that x bar is equal to 1 over m times the integral of xp of x, y, z, d, s.

00:22 And likewise, the same is true for y bar and z bar.

00:26 They are all equal to 1 over m, multiplied by the integral of s, except.

00:34 It will be y and z instead of x.

00:40 So p, p of x, y, z, d, s of x, y, z, d s, and m, m is equal to the surface integral of p of x, y, z, d, s.

01:01 So we can rewrite the function given as z is equal to the square root of a squared minus x squared minus y squared.

01:15 And we know that ds is equal to when you can write, when you can write z in terms of x, y.

01:23 We know that ds is equal to the square root of one plus the partial with respect to x squared plus the partial with respect to y squared, d, y, d, x.

01:38 That's equal to d s.

01:40 And so in this case, ds will be equal to a over the square root of a squared minus x squared minus y squared, d, y, dx.

02:02 And p in this case represents the density, by the way, and p is a constant.

02:07 It's just important because we can always pull it outside for the integrals.

02:13 So, d in this case, for the integral, the integral of d, p of a over square root of a squared minus x squared plus y squared, d, y, d, x, in this case, is a disk with radius a, and it's centered at the origin.

02:37 So in polar coordinates, this is equal to the integral of p, a .r, or par, of square root, a squared minus r squared, d, d, d, theta...

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