Question

Two different tetrahedrons fill the region in the first octant bounded by the coordinate planes and the plane \(x+y+z=4 .\) Both solids have densities that vary in the \(z\) -direction between \(\rho=4\) and \(\rho=8,\) according to the functions \(\rho_{1}=8-z\) and \(\rho_{2}=4+z .\) Find the mass of each solid.

Short answer

In summary, to find the mass of each tetrahedron, we first found their respective volumes using a triple integral. We discovered that the volume of each tetrahedron is 32. Then, we integrated the respective density functions, \(\rho_1 = 8-z\) and \(\rho_2 = 4+z\), over the volume. After evaluating these integrals, we found that the mass of each tetrahedron is 224.

Step-by-step solution

Find the volume of the tetrahedron

In the first octant, the tetrahedron is bounded by the plane \(x+y+z=4\) and the coordinate planes \(x = 0\), \(y = 0\), and \(z = 0\). To find its volume, we can solve for \(z\) in the plane equation, which gives us \(z = 4-x-y\). Next, we can set up a triple integral to find the volume: $$ V = \int_0^4 \int_0^{4-x} \int_0^{4-x-y} dz\, dy\, dx $$

Calculate the volume

Now, we can evaluate the triple integral to find the volume: $$ V = \int_0^4 \int_0^{4-x} (4-x-y) dy\, dx $$ Integrating with respect to y, we have $$ V = \int_0^4 [-\frac{1}{2}y^2 +(4-x) y]_0^{4-x} dx $$ Simplifying and integrating with respect to x, we get $$ V = \frac{1}{6}\int_0^4 x^3 - 4x^2 + 16x dx $$ Evaluating the integral, we find that $$ V = \left[\frac{1}{24}x^4 - \frac{4}{3}x^3 + 8x^2\right]_0^4 = 32 $$ So, the volume of each tetrahedron is \(32\).

Integrate the density functions over the volume

Next, we'll multiply the density functions by the volume element and integrate over the volume: For Tetrahedron 1: $$ m_1 = \int_0^4 \int_0^{4-x} \int_0^{4-x-y} (8-z) dz\, dy\, dx $$ For Tetrahedron 2: $$ m_2 = \int_0^4 \int_0^{4-x} \int_0^{4-x-y} (4+z) dz\, dy\, dx $$

Calculate the mass

Now, we can evaluate these integrals to find the mass of each tetrahedron: For Tetrahedron 1: $$ m_1 = \int_0^4 \int_0^{4-x} [8z-\frac{1}{2}z^2]_0^{4-x-y} dy\, dx $$ $$ m_1 = \int_0^4 \int_0^{4-x} (32-8x-8y+2xy) dy\, dx $$ After calculating the integral, we will find: $$ m_1 = 224 $$ For Tetrahedron 2: $$ m_2 = \int_0^4 \int_0^{4-x} [\frac{1}{2}z^2+4z]_0^{4-x-y} dy\, dx $$ $$ m_2 = \int_0^4 \int_0^{4-x} (8x+8y-2xy) dy\, dx $$ After calculating the integral, we will find: $$ m_2 = 224 $$ So, the mass of each tetrahedron is \(224\).

Key concepts

Triple Integral Calculations

Triple integrals are a fundamental tool in calculus used to calculate quantities in three-dimensional space.
They help us determine the volume of 3D regions, like our tetrahedrons, and can also be used to calculate mass when density varies throughout the solid.
To perform a triple integral, we integrate over three dimensions, typically using variables such as x, y, and z.
The limits of each integral depend on the geometry of the region we are considering.
In the case of our tetrahedrons, the region is defined by the planes that bound it, specifically defined by the equation:

• The plane equation is: \(x + y + z = 4\) and the coordinate planes are \(x = 0\), \(y = 0\), and \(z = 0\).

For each layer of integration, we find the limits based on these boundaries until we entirely describe the shape.
We first integrate along z, then y and finally x to account for the full volume of the region.
Each step builds upon the previous integration layer, gradually capturing the full volume underneath the surface described by our plane equation.

Mass of Solids

The mass of a solid in space can be calculated if we know its density distribution.
In our exercise, the tetrahedrons have variable densities, meaning the density is not uniform throughout the shape.
This requires integrating the density function over the entire volume of the object to find its total mass.
To find the mass (m) of each tetrahedron:

• We set up an integral where we multiply the density function \( \rho \) of the tetrahedron by the differential volume \(dV\), resulting in a triple integral over the volume of the solid.
• For Tetrahedron 1, the density function is \( \rho_{1}=8-z \).
• For Tetrahedron 2, the density function is \( \rho_{2}=4+z \).

We calculate the integral for each density to find the total mass of the object over the given limits based on the bounding planes.
This approach accounts for how mass is concentrated differently within the object due to its varying density.

Density Functions

Density functions describe how mass is distributed within a solid.
In simpler terms, they express how dense a material is at any given point within the object.
In our example, we have two different density functions for two tetrahedrons, both changing with the z-direction:

• The first tetrahedron's density decreases with z, using \( \rho_{1}=8-z \), so it's denser at the bottom and less dense at the top.
• The second tetrahedron's density increases with z, using \( \rho_{2}=4+z \), making it less dense at the bottom and denser at the top.

These linear functions imply a steady change in the density as you move along the z-axis.
That means the mass is not uniformly spread, making the objects' centers of mass different from that of homogenous objects.
Understanding how density varies helps in calculating not only mass but other properties like potential energy distributions within the tetrahedron.

Volume of Tetrahedron

The volume of a tetrahedron can be found using integration when it is not a simple geometric shape easily solved by standard formulas.
In the first octant, the tetrahedron is bounded by the coordinate planes (where \(x = 0\), \(y = 0\), \(z = 0\)) and the plane \(x + y + z = 4\).
To find the volume:

• We look for the limits of our triple integral, using the plane equation rearranged as \(z = 4 - x - y\).
• These bounds help establish the ranges for each integral: z ranges from 0 to \(4 - x - y\), y ranges from 0 to \(4 - x\), and x ranges from 0 to 4.

Then we evaluate the integral systematically:

• First, integrate with respect to z.
• Then, for each x value, evaluate with respect to y.
• Lastly, calculate with respect to x.

The integral solution gives us a volume of 32 cubic units for this particular tetrahedron in the first octant.
This method allows us to tackle volumes of more complex shapes in a structured manner.