Question

Use double integrals to calculate the volume of the following regions. The tetrahedron bounded by the coordinate planes \((x=0, y=0, z=0)\) and the plane \(z=8-2 x-4 y\)

Short answer

Answer: The volume of the tetrahedron is 0 cubic units.

Step-by-step solution

Finding the vertices of the tetrahedron

The tetrahedron is formed by the intersection of the plane \(z=8-2x-4y\) and the coordinate planes \(x=0\), \(y=0\), \(z=0\). We can find the three vertices by setting two of the three variables (x, y, and z) to 0 and solving for the remaining variable. 1. x = 0, y = 0: \(z = 8 - 2(0) - 4(0) = 8\), so the first vertex is (0, 0, 8). 2. x = 0, z = 0: \(0 = 8 - 2(0) - 4y\); \(y = 2\), so the second vertex is (0, 2, 0). 3. y = 0, z = 0: \(0 = 8 - 2x - 4(0)\); \(x = 4\), so the third vertex is (4, 0, 0).

Setting up the double integral

We will use the double integral over the xy-plane for the height function z, where z = 8 - 2x - 4y. From the vertices, we know that x is bounded by [0, 4] and y is bounded by [0, 2]. Therefore, the double integral is given by: $$\int_{0}^{4} \int_{0}^{2} (8-2x-4y)\,dy\,dx$$

Evaluating the integral

First, we'll integrate with respect to y: $$ \int_{0}^{4} \Biggl[ \left(8y - 2xy - 2y^2\right) \Biggr|_0^2\,dx $$ Next, evaluate the limits of integration for y: $$ \int_{0}^{4} (16 - 4x - 8)\,dx $$ Simplify: $$ \int_{0}^{4} (8 - 4x)\,dx $$ Now, integrate with respect to x: $$ \Biggl[8x - 2x^2 \Biggr|_0^4 $$ Evaluate the limits of integration for x: $$ (8 \cdot 4 - 2 \cdot 4^2) - (0) = 32 - 32 = 0 $$ Finally, the volume of the tetrahedron is 0 cubic units.

Key concepts

Volume Calculation

Calculating volumes using double integrals involves integrating a function over a specific region. For a solid lying below a surface, the function typically represents the height above each point in the region. To find the volume of our tetrahedron, we choose the region in the xy-plane under the curve described by the plane equation, which is given by its height function. This height function, or surface equation, is expressed as \(z = 8 - 2x - 4y\). To compute the volume, we integrate this function over the bounds that describe the triangular region where the base of our tetrahedron lies. This application of double integrals becomes a powerful tool for finding volumes in three dimensions.

Tetrahedron

A tetrahedron is a simple three-dimensional shape with four triangular faces, which can be visualized as a pyramid with a triangular base. In this case, our tetrahedron is defined by the coordinate planes and a given plane equation \(z = 8 - 2x - 4y\). To understand its shape, we determine where the plane intersects the coordinate axes, resulting in vertices at \((0, 0, 8)\), \((0, 2, 0)\), and \((4, 0, 0)\). These points mark the corners of the base and apex of the pyramid. The coordinate planes \(x=0\), \(y=0\), and \(z=0\) essentially carve out the solid region of interest from the infinite space defined by the abovementioned plane.

Coordinate Planes

The coordinate planes are fundamental references in three-dimensional space, setting the boundaries of the regions of interest in many problems. For the tetrahedron problem, these include the xy-plane, yz-plane, and xz-plane. Each plane corresponds to setting one of the coordinates \(x\), \(y\), or \(z\) to zero respectively. In our scenario, they represent the faces of the tetrahedron meeting at the origin. By exploring where these planes intersect with our given plane \(z = 8 - 2x - 4y\), we find the vertices of the tetrahedron, essentially carving and defining the triangular slice of space within the first octant.

Integration Bounds

Integration bounds define the limits within which the integration takes place over a region. In a double integral, these bounds typically cover the other two dimensions orthogonally depending on the axes considered. For the tetrahedron in question, bounds are set in the \(x\) and \(y\) dimensions. The \(x\) limits range from 0 to 4, and for each \(x\) value, \(y\) spans from 0 to a linear limit defined by the intersection with the plane, which is actually a line equation derived from the triangular base. These bounds ensure the integral describes the full volume of the tetrahedron between the plane and the coordinate planes. By integrating the height function \(z = 8 - 2x - 4y\) over these specified bounds, we fully capture the desired volume within the defined region.