Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 21-32, sketch the indicated solid. Then find its volume by an iterated integration. Tetrahedron bounded by the coordinate planes and the plane \(z=6-2 x-3 y\)

Step-by-Step Solution

Answer

The volume of the tetrahedron is 6 cubic units.

Step 1: Understand the Problem

We need to find the volume of a tetrahedron which is formed by the intersection of the plane \( z = 6 - 2x - 3y \) with the coordinate planes \( x = 0 \), \( y = 0 \), and \( z = 0 \). This solid is bounded in the first octant.

Step 2: Find the Intercepts

Determine where the plane \( z = 6 - 2x - 3y \) intersects the coordinate axes. Set \( y = 0 \) and \( z = 0 \) to find the x-intercept: \( 6 - 2x = 0 \Rightarrow x = 3 \). Similarly, set \( x = 0 \) and \( z = 0 \) to find the y-intercept: \( 6 - 3y = 0 \Rightarrow y = 2 \). For the z-intercept, set \( x = 0 \) and \( y = 0 \) to find \( z = 6 \).

Step 3: Set Up the Double Integral

To find the volume, we set up the double integral based on the region of integration in the \( xy \)-plane. It is a triangular region with vertices at \((0,0)\), \((3,0)\), and \((0,2)\). We use \( y \) as the inner integral bounds: \( 0 \leq y \leq 2 \), and \( x \) as the outer integral bounds: \( 0 \leq x \leq 3 - \frac{3}{2}y \). The integrand is the function \( z = 6 - 2x - 3y \).

Step 4: Compute the Inner Integral

Calculate the inner integral with respect to \( y \):\[ \int_0^{3 - \frac{3}{2}y} (6 - 2x - 3y) \, dx \].Solve the inner integral:\[ \int (6 - 2x - 3y) \, dx = 6x - x^2 - 3xy \] evaluated from \( x = 0 \) to \( x = 3 - \frac{3}{2}y \).

Step 5: Evaluate the Inner Integral

Substitute the limits into the evaluated indefinite integral: \[ (6(3 - \frac{3}{2}y) - (3 - \frac{3}{2}y)^2 - 3y(3 - \frac{3}{2}y)) - (6(0) - (0)^2 - 3(0)y) \].Simplify the expression to get:\[ 18 - 9y - (9 - 9y + \frac{9}{4}y^2) + 9y - \frac{9}{2}y^2 = \frac{9}{4}y^2.\]

Step 6: Compute the Outer Integral

Now integrate with respect to \( y \):\[ \int_0^2 \frac{9}{4}y^2 \, dy. \]This becomes \[ \frac{9}{4} \cdot \frac{y^3}{3} \] evaluated from \( y = 0 \) to \( y = 2 \).

Step 7: Evaluate the Outer Integral

Substitute \( y = 2 \) and \( y = 0 \) into the antiderivative:\[ \frac{9}{4} \times \frac{8}{3} = 6. \]Thus, the volume of the tetrahedron is 6 cubic units.

Key Concepts

Iterated Integration

Iterated integration is a method used to compute the volume of a solid by integrating over its boundary in a step-by-step manner. This technique breaks down a complex integral into simpler, sequential steps, making it easier to evaluate. In the case of the given tetrahedron, we use iterated integration to first integrate with respect to one coordinate, and then integrate the resulting expression with respect to another coordinate.
By setting up the integral for each dimension in the order dictated by the solid's geometry, we simplify the computation process. This approach is particularly useful when we deal with rectangular or triangular regions in a coordinate plane.

• Start by determining the bounds for the innermost integral, which corresponds to one of the variables.
• Evaluate this integral to simplify to a function of the remaining variables.
• Continue this process, integrating each remaining variable in sequence, until you arrive at a numerical answer representing the volume.

This method efficiently handles problems involving multiple variables and complex limits, such as the volume of a tetrahedron within specific boundaries.

Coordinate Geometry

Coordinate geometry provides a systematic method to analyze spatial relationships between points, lines, and shapes using a coordinate plane. In solving for the volume of a tetrahedron bounded by coordinate planes and a given plane, we rely heavily on coordinate geometry to understand its dimensions and limits.
The given equation \( z = 6 - 2x - 3y \) represents a plane, which intersects with the coordinate planes \( x = 0 \), \( y = 0 \), and \( z = 0 \) to form a tetrahedron. Using coordinate geometry, we can determine the intercepts, which are key points where the plane meets the coordinate axes. These intercepts serve as vertices of the triangular base of the tetrahedron within the three-dimensional coordinate system.

• The x-intercept is found by setting \( y = 0 \) and \( z = 0 \). From the equation, \( x = 3 \).
• The y-intercept is obtained by setting \( x = 0 \) and \( z = 0 \), giving us \( y = 2 \).
• The z-intercept, where both \( x \) and \( y \) are 0, is \( z = 6 \).

These intercepts provide crucial reference points for sketching and analyzing the tetrahedron, ultimately guiding us in setting up the necessary integrals.

Triple Integrals

Triple integrals extend the concept of ordinary integration to three dimensions, allowing us to calculate volumes of solids bounded by specific surfaces. These integrals are an essential tool in multivariable calculus and come into play when analyzing three-dimensional shapes, such as the given tetrahedron.
In our problem, the integration starts with setting up a double integral over the base, followed by an integral over the height of the solid to arrive at the final volume.

• Each integral handles a different dimension (or direction) of the space containing the solid.
• We apply the triple integral concept by solving two iterated integrals: one for the area and one for the height.
• This approach ensures the calculation of the entire region defined by the given boundaries.

Essentially, working through triple integrals involves stacking each subsequent layer of integration atop the previous one, resulting in a comprehensive calculation of volume.

Solid Boundaries

Solid boundaries define the physical constraints or edges within which a solid exists. Understanding these boundaries is crucial for setting up the integration limits accurately. For the tetrahedron in question, these boundaries are determined by the intersecting planes and the coordinate planes in the first octant.
The boundaries for the region in the xy-plane are particularly important. In the case of our tetrahedron, the boundaries form a triangular region bounded by the lines:

• \( x = 0 \)
• \( y = 0 \)
• \( x = 3 - \frac{3}{2}y \)

The intersections define the spatial confines of the region that must be considered during the integration process. Properly establishing these limits is fundamental to accurately computing the volume, ensuring that the computed integral represents only the enclosed space of interest, and not beyond.Exploring these boundaries allows us to visualize and sketch the solid before involving ourselves in complex integrations, providing a clearer path to the solution.