Question

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and h are positive constants. Find the volume of a tetrahedron whose vertices are located at \((0,0,0),(a, 0,0),(0, b, 0),\) and \((0,0, c)\)

Short answer

Answer: The volume of the tetrahedron is \(\frac{abc}{6}\).

Step-by-step solution

Equations for the bounding surfaces

We need to find the equations for the four planes that define the tetrahedron. These planes are formed by the intersection of the planes containing the three axes (x, y, and z) and the plane passing through the three vertices (a, 0, 0), (0, b, 0), and (0, 0, c). The equations of the planes are as follows: Plane 1: x = 0 (yz-plane) Plane 2: y = 0 (xz-plane) Plane 3: z = 0 (xy-plane) Plane 4: To find the equation of the fourth plane, we can use the determinant formula: | x y z 1 | | a 0 0 1 | | 0 b 0 1 | | 0 0 c 1 | Expanding this determinant along the first row gives us the equation of the fourth plane, which is: Plane 4: \(bcx + acy + abz - abc = 0\)

Set up a volume integral

To find the volume of the tetrahedron, we will set up a triple integral for the region enclosed by these planes. The integral is bounded by the four planes we found in step 1. \(V = \int\int\int_{V} dV\) The bounds for the integral are as follows: For x: 0 to a (due to the xz-plane) For y: 0 to \(b \left(1 - \frac{x}{a}\right)\) (due to the intersection of the xy-plane and Plane 4) For z: 0 to \(c \left(1 - \frac{x}{a} - \frac{y}{b}\right)\) (due to the intersection of the yz-plane and Plane 4) Putting this together, the integral becomes: \(V = \int_{0}^{a} \int_{0}^{b\left(1 - \frac{x}{a}\right)} \int_{0}^{c\left(1 - \frac{x}{a} - \frac{y}{b}\right)} dz\, dy\, dx\)

Evaluate the integral

Now, we will evaluate the integral to find the volume of the tetrahedron: \(V = \int_{0}^{a} \int_{0}^{b\left(1 - \frac{x}{a}\right)} \left[c\left(1 - \frac{x}{a} - \frac{y}{b}\right)\right] dy\, dx\) Evaluate the inner integral with respect to y: \(V = \int_{0}^{a} \left[ -\frac{bcy^2}{2b} + \frac{bcy^{3}}{3b^2}\right]_{0}^{b\left(1 - \frac{x}{a}\right)} dx\) \(V = \int_{0}^{a} \left[ -\frac{bc\left(b\left(1 - \frac{x}{a}\right)\right)^2}{2} + \frac{bc\left(b\left(1 - \frac{x}{a}\right)\right)^{3}}{3b}\right] dx\) Simplify and evaluate the outer integral with respect to x: \(V = \frac{abc}{6} \int_{0}^{a} \left[ 6 - 6\frac{x}{a} - 3\frac{x^2}{a^2}\right] dx\) \(V = \frac{abc}{6} \left[6x - 3\frac{x^2}{a} - \frac{x^3}{a^2}\right]_{0}^{a}\) \(V = \frac{abc}{6} \left[6a - 3a + a\right]\) Finally, we obtain the volume of the tetrahedron: \(V = \frac{abc}{6}\) Hence, the volume of the tetrahedron with vertices at (0,0,0), (a,0,0), (0,b,0), and (0,0,c) is given by \(\frac{abc}{6}\).

Key concepts

Bounding Surfaces

In the context of geometry and calculus, bounding surfaces are crucial for defining the limits of a three-dimensional region like a tetrahedron. A tetrahedron is a solid shape with four triangular faces, and in this exercise, it is defined by four planes. These planes are mathematical expressions that create the confines for the shape. The three simpler planes correspond to the coordinate planes:

• The yz-plane is defined by x = 0
• The xz-plane by y = 0
• The xy-plane by z = 0

To complete the tetrahedron, a fourth plane is formed using the vertices of the shape, given as (a, 0, 0), (0, b, 0), and (0, 0, c). The equation of this plane can be determined using a determinant formula, which is essentially a mathematical tool that helps derive the relationship among the plane's points.The fourth plane's equation is crafted using the formula of the determinant. This results in an equation that connects x, y, and z in relation to the constants a, b, and c: \(bcx + acy + abz - abc = 0\). This equation ensures that any point (x, y, z) on this plane will satisfy this relation, thus forming the fourth and final bounding surface of the tetrahedron.

Triple Integral

Triple integrals are indispensable in calculating volumes of three-dimensional spaces enclosed by surfaces like our tetrahedron. It's like stacking layers of infinitesimally thin sheets over the entire region. Using triple integrals, we integrate over three variables, which in this example are x, y, and z.For the given tetrahedron, we first identify the boundaries of these variables based on the bounding planes found earlier:

• The x-values range between 0 and a, guiding the outermost boundary.
• The y-values depend on x and range from 0 to \(b\left(1 - \frac{x}{a}\right)\).
• The z-values hinge on both x and y, ranging from 0 to \(c\left(1 - \frac{x}{a} - \frac{y}{b}\right)\).

Thus, the entire volume is evaluated through a triple integral: \[ V = \int_{0}^{a} \int_{0}^{b\left(1 - \frac{x}{a}\right)} \int_{0}^{c\left(1 - \frac{x}{a} - \frac{y}{b}\right)} dz\, dy\, dx \]This expression stacks together all the infinitesimal elements \(dz\, dy\, dx\) that collectively fill up the volume of the tetrahedron.

Determinant Formula

The determinant formula plays a vital role in geometry for finding equations of planes and computing areas and volumes. In our exercise, it is used to find the fourth plane of the tetrahedron, which is essential to its volume calculation. To determine the relationship connecting the points of the plane, we use the determinant of a 4x4 matrix. This matrix includes the coordinates of the points (a, 0, 0), (0, b, 0), and (0, 0, c), with an additional row that accounts for homogeneous coordinates, ensuring all points lay on the plane. The formula looks like this:\[\left|\begin{array}{cccc} x & y & z & 1 \ a & 0 & 0 & 1 \ 0 & b & 0 & 1 \ 0 & 0 & c & 1\end{array}\right|\]By expanding this determinant, we derive the equation \(bcx + acy + abz - abc = 0\). This equation represents a linear combination of x, y, and z, establishing the necessary bounds for determining the overall volume of the tetrahedron. Understanding and applying this formula are key for ensuring accurate geometric and volumetric calculations.