Chapter 5: Problem 47

Find the volume in the first octant bounded by the coordinate planes and the planc \(x+2 y+z=4\)

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Understand the Boundaries

Identify the volume confined by the coordinate planes (the planes at x=0, y=0, and z=0) and the plane given by the equation \(x + 2y + z = 4\). Since we are in the first octant, all coordinates (x, y, and z) must be non-negative.

Set Up the Triple Integral

To find the volume, set up a triple integral \[ V = \int_{0}^{2} \int_{0}^{2-\frac{x}{2}} \int_{0}^{4-x-2y} dz\ dy\ dx \]. Identify the limits of integration based on the intersections:

Evaluate the Inner Integral

Evaluate the inner integral with respect to z:\[ \int_{0}^{4-x-2y} dz = [z]_{0}^{4-x-2y} = (4-x-2y) - 0 = 4 - x - 2y\]

Evaluate the Second Integral

Now evaluate the integral with respect to y:\[ \int_{0}^{2-\frac{x}{2}} (4 - x - 2y) dy \] Compute this integral step:\[ \int_{0}^{2-\frac{x}{2}} (4-x) dy - \int_{0}^{2-\frac{x}{2}} 2y dy \ \int_{0}^{2-\frac{x}{2}} (4-x) dy = (4-x)y \] evaluated from 0 to (\(2-\frac{x}{2}\)):\[ (4-x)(2-\frac{x}{2}) \]\[ = 8-2x-x^2/2 \]\[ \int_{0}^{2-\frac{x}{2}} 2y dy = y^2 \] evaluated from 0 to (\(2-\frac{x}{2}\)):\[ y^2 \]\[ = 4-2x+x^2/4 \]All combinations:\[ = 8-2x-x^2/2-4+2x-x^2/48-4-3x^2/4 \]

Evaluate the Outer Integral

Finally, evaluate the last integral with respect to x:\[ \int_{0}^{4-x-2y}(4-x) dy - \]\[.... \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Calculation

Volume calculation using triple integrals is a powerful tool in mathematics. It helps us find the volume of complex shapes in 3D space. In this context, we will break down the process to ensure it is easily understandable. To start, we employ the triple integral, which is the integration of a function of three variables over a three-dimensional region.

For our problem, we set up the integral to find the volume within the first octant, bounded by the plane \(x + 2y + z = 4\) and the coordinate planes. This requires an \(dx, dy, dz\) integral setup to cover all boundaries. Setting the correct limits involves understanding how the plane intersects the axes:

At \(x=0\), solving \(2y + z = 4\).
At \(y=0\), solving \(x + z = 4\).
At \(z=0\), solving \(x + 2y = 4\).

By solving these, we find the limits of integration for \(x\, y,\,\ z\), then integrate step-by-step to compute the volume effectively.

First Octant

The first octant in three-dimensional space is where all three coordinates (x, y, and z) are non-negative. It lies between the positive sides of the coordinate planes: the \(xy\) plane, the \(xz\) plane, and the \(yz\) plane.

When finding the volume in the first octant, it simplifies our calculations as we don't have to consider negative values for x, y, or z. For our problem, this confines the volume under the given plane and within positive x, y, and z limits.

Thus, our integrations naturally go from \(0\) to the respective intersections (or boundaries) derived from the plane equation. This narrows down our work and ensures all values considered are within this first octant zone.

Coordinate Planes

Coordinate planes are the \(xy\), \(xz\), and \(yz\) planes in three-dimensional space. They are essential in defining boundaries for volume calculations.

The \(xy\) plane is where \(z=0\). It spans the first quadrant of the \(x\) and \(y\) axes. Similarly, the \(xz\) plane (where \(y=0\)) and the \(yz\) plane (where \(x=0\)) provide other boundary conditions.

The \(xy\) plane helps in integrating z from 0 to its boundary value.
The \(xz\) plane handles y values.
The \(yz\) plane manages x values.

For the given plane \(x + 2y + z = 4\), it interacts with all three coordinate planes, helping us derive the integral bounds. Recognizing these planes is critical as they simplify our integral setup and ensure accurate computation within the predefined limits.