Centroids in 3D via the First Moment Integral

The centroid of a volume can be thought of as the geometric center of that area. It is the average position (x, y, and z coordinates) of all the points in in the area. If this volume represents a part with a uniform density (like most single material parts) then the centroid will be the same as the center of mass.

Centroids are useful for a number of situations in the mechanics course sequence, including the analysis of distributed forces, determining the center of mass, and as an intermediate step in determining moments of inertia which are used to determining an object's resisting to rotational accelerations or an objects resistance to bending or torsion.

Just as we had with centroids of areas, two methods exist for finding centroids of volumes:

  1. Using calculus and the first moment integral.
  2. Using the method of composite parts and tables of centroids for common volumes.

The tables used in the method of composite parts however are derived from the first method, so both methods ultimately use the moment integrals.

Finding the Centroid via the First Moment Integral

When we find the centroid of a three dimensional shape, we will be looking for x, y, and z coordinates (x bar, y bar, and z bar). This will be the x, y, and z coordinates of the point that is the centroid of the shape.

Much like the centroid calculations we did with 2D shapes, we are looking to find the shapes average coordinate in each dimension. We do this by summing up all the little bits of volume times the x, y, or z coordinate of that bit of volume and then dividing that sum by the total volume of the shape. Again we will use calculus to sum up an infinite number of infinitely small volumes. Specifically this sum will be the first, rectangular, volume moment integral for the shape. Working in each of the three coordinate directions we wind up with the following three equations.

Centroid Equation 3D

With these new equations we have the variable dV rather than dA, since we are integrating over a volume rather than an area. This represents the rate of change of the volume as we move along an axis from one end to another. The rate of change of the volume at any point on the shape will be the cross sectional area that is perpendicular to that axis. Since cross sectional area varies as we move along the axis, we will need to determine a formula for the cross sectional area at any point along that axis.

The center of mass
The z coordinate of the center of mass.

Using the first moment integral and the equations shown above we can theoretically find the centroid of any volume as long as we can write an equation to describe the cross section area at for each direction. For more complex shapes however, determining these equations and then integrating these equations may become very time consuming. For these complex shapes, the method of composite parts or computer tools will most likely be much faster.

Using Symmetry as a Shortcut

Just as with 2D areas, shape symmetry can provide a shortcut in many centroid calculations. Remember that the centroid coordinate is the average x, y, and z coordinate for all the points in the shape. If the volume has a plane of symmetry, that means each point on one side of the line must have an equivalent point on the other side of the line. This means that the average value (aka the centroid) must lie within that plane. If the volume has more than one plane of symmetry, then the centroid must exist at the intersection of these planes.

Centroids and Symmetry
If a volume has a plane of symmetry, the centroid must lie somewhere in that plane. If the volume has more than one plane of symmetry, the centroid must exist at the intersection of these planes.

Worked Problems:

Question 1:

The cone shown below is four inches tall and has a four inch diameter base. Find the x, y, and z coordinates of the centroid.

Problem 1 Diagram

Solution:



Question 2:

Find the y coordinate of the centroid for the tetrahedron shown in the image below. (The fourth vertex is at the origin)

Problem 2 Diagram

Solution: