Question

Find the center of mass of the following solids, assuming a constant density of 1. Sketch the region and indicate the location of the centroid. Use symmetry when possible and choose a convenient coordinate system. The tetrahedron in the first octant bounded by \(z=1-x-y\) and the coordinate planes

Short answer

Answer: The centroid of the tetrahedron is located at the point (1/4, 1/4, 1/4).

Step-by-step solution

Define the region of integration

First, let's determine the limits of integration for the x, y, and z coordinates. We are working in the first octant, which means x, y, and z will all be non-negative (greater than or equal to 0). The surfaces bounding the tetrahedron region are given by x = 0, y = 0, z = 0, and z = 1 - x - y.

Find the mass of the region

To find the mass of the region, we need to integrate the density function (which is a constant 1) over the region and obtain the total mass. The mass can be computed as: $$M = \int\int\int_D 1 \,dV$$ where D is the region of the tetrahedron.

Calculate the centroid coordinates

To find the centroid coordinates, we need to compute the three moments, M_x, M_y, and M_z, and then divide each of them by the mass M. We can calculate these moments using triple integrals over the region. $$M_x = \int\int\int_D y \,dV$$ $$M_y = \int\int\int_D x \,dV$$ $$M_z = \int\int\int_D z \,dV$$ Let's compute these moments and the mass M:

Set up the triple integrals

We have already defined the limits of integration for the x, y, and z-coordinates, so now we need to set up the triple integrals for the mass and moments. Since we are working in Cartesian coordinates, we needs to keep the order of integration in mind. We can integrate with respect to z first, then y, and finally x: $$M = \int_0^1 \int_0^{1 - x} \int_0^{1 - x - y} 1 \,dz\,dy\,dx$$ $$M_x = \int_0^1 \int_0^{1 - x} \int_0^{1 - x - y} y \,dz\,dy\,dx$$ $$M_y = \int_0^1 \int_0^{1 - x} \int_0^{1 - x - y} x \,dz\,dy\,dx$$ $$M_z = \int_0^1 \int_0^{1 - x} \int_0^{1 - x - y} z \,dz\,dy\,dx$$

Evaluate the triple integrals

To compute the centroid coordinates, let's first evaluate all the triple integrals. These are standard integrals that can be calculated using the Generalized Fundamental Theorem of Calculus. The results are: $$M = \frac{1}{6}$$ $$M_x = \frac{1}{24}$$ $$M_y = \frac{1}{24}$$ $$M_z = \frac{1}{24}$$

Find the centroid coordinates

Now, let's find the coordinates of the centroid (center of mass) by dividing the moments by the mass (x̄ = M_x/M, ȳ = M_y/M, and ẑ = M_z/M): $$x̄ = \frac{M_x}{M} = \frac{\frac{1}{24}}{\frac{1}{6}} = \frac{1}{4}$$ $$ȳ = \frac{M_y}{M} = \frac{\frac{1}{24}}{\frac{1}{6}} = \frac{1}{4}$$ $$ẑ = \frac{M_z}{M} = \frac{\frac{1}{24}}{\frac{1}{6}} = \frac{1}{4}$$ So the centroid coordinates are (x̄, ȳ, ẑ) = (1/4, 1/4, 1/4)

Sketch the region and the centroid

Now, let's draw the tetrahedron region using the given surfaces (z = 1 - x - y, x = 0, y = 0, and z = 0) and mark the position of the centroid (1/4, 1/4, 1/4), with respect to the x, y and z axes.  The centroid (center of mass) of this tetrahedron is located at the point (1/4, 1/4, 1/4).

Key concepts

Tetrahedron

A tetrahedron is a three-dimensional geometric shape with four triangular faces, six edges, and four vertices. Imagine a pyramid with a triangular base, and you have a basic image of a tetrahedron. In this context, we are looking at a special type known as a right tetrahedron formed in the first octant of a coordinate system.

The first octant is the section of the three-dimensional Cartesian coordinate system where all coordinates

• x, y, and z are positive.

The given problem involves a tetrahedron bounded by the coordinate planes x = 0, y = 0, z = 0, and the plane defined by the equation z = 1 - x - y. This plane intersects the axes at x = 1, y = 1, and z = 1, forming the triangular faces.

This tetrahedron's base lies flat on the xy-plane, and its apex points upwards towards the z-axis, providing a useful visualization before any calculations.

Triple Integrals

Triple integrals are a mathematical tool used to calculate volume and other quantities, such as mass or centroid coordinates, in three-dimensional space. They extend the concept of double integrals, which compute area, by adding an additional dimension. A triple integral involves integrating a function over a three-dimensional region.

In this exercise, we use triple integrals to find the mass and moments of inertia for the tetrahedron. Since the density is constant and equal to 1, the integral for mass simplifies greatly:

• Mass is calculated by integrating 1 over the region D.

The triple integrals take the form:\[ M = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} 1 \,dz\,dy\,dx \]Here, the limits of integration correspond to the boundaries of the tetrahedron within the first octant.

Once you set up the integral, you evaluate it by integrating with respect to z first, then y, and finally x. This ordered integration simplifies computations and aligns with the boundaries defined by z = 1 - x - y.

Centroid Calculation

The centroid, also known as the center of mass, is a point that acts as if all the mass of the object is concentrated there. For uniform density objects, it's the average position of all the points in the space.

To find a tetrahedron's centroid, you calculate the moments of the region with respect to the coordinate planes. This is done by integrating functions representing the respective coordinate variables over the region:

• \(M_x\): moment with respect to the yz-plane through \(\int\int\int y \,dV\)
• \(M_y\): moment with respect to the xz-plane through \(\int\int\int x \,dV\)
• \(M_z\): moment with respect to the xy-plane through \(\int\int\int z \,dV\)

After computing the integrals for these moments, find the centroid coordinates \((\bar{x}, \bar{y}, \bar{z})\) by dividing each moment by the total mass.
For instance, the centroid coordinates \((\bar{x}, \bar{y}, \bar{z})\) in this problem are calculated as \((1/4, 1/4, 1/4)\), meaning the point is at one fourth along each axis from the origin.

Coordinate System

The coordinate system is a framework used to locate points in space. In this exercise, we use a three-dimensional Cartesian coordinate system. It consists of mutually perpendicular axes: x, y, and z, intersecting at an origin point.

Each point within this system is defined by an ordered triplet of numbers \((x, y, z)\), representing distances along these axes from the origin. This system allows precise location of the tetrahedron's vertices, faces, and facilitates the integration process.

• The first octant, where this particular tetrahedron is located, includes all points where x, y, and z are positive.
• The boundary planes x = 0, y = 0, and z = 0 are part of the coordinate planes that confine the shape within this octant.

Choosing a convenient coordinate system, like the Cartesian one used in this problem, simplifies integration and visualization, making it easier to solve for the centroid effectively and efficiently.