/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 \(\iiint_{T} x y z d V,\) where ... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 14

\(\iiint_{T} x y z d V,\) where \(T\) is the solid tetrahedron with vertices \((0,0,0),(1,0,0),(1,1,0),\) and \((1,0,1)\)

Short Answer

Expert verified
The integral evaluates to \( \frac{1}{140} \).

Step by step solution

01

Determine the Boundaries of the Integral

The solid tetrahedron is bounded by the planes represented by the vertices: 1. The base on the xy-plane is given by the vertices (0,0,0), (1,0,0), (1,1,0). Thus, the limits for x and y over that base are 0 鈮� x 鈮� 1 and 0 鈮� y 鈮� x. 2. The height in the z-direction is from z = 0 to the plane z = 1 - x. Thus, z is bounded by 0 鈮� z 鈮� 1 - x. This gives the nested limits of integration: 0 鈮� z 鈮� 1 - x, 0 鈮� y 鈮� x, 0 鈮� x 鈮� 1.
02

Set Up the Triple Integral

Based on the limits identified, the triple integral becomes: \[ \int_{0}^{1} \int_{0}^{x} \int_{0}^{1-x} x y z \, dz \, dy \, dx \] This sets up the sequential integration for z, then y, and finally x.
03

Perform the Integration with Respect to z

First, integrate with respect to z: \[ \int_{0}^{1-x} x y z \, dz = xy \left[ \frac{z^2}{2} \right]_{0}^{1-x} = xy \left( \frac{(1-x)^2}{2} \right) \] Simplifying gives: \[ \frac{xy(1-x)^2}{2} \]
04

Perform the Integration with Respect to y

Integrate the result with respect to y:\[ \int_{0}^{x} \frac{xy(1-x)^2}{2} \, dy = \frac{(1-x)^2}{2} \left[ \frac{xy^2}{2} \right]_{0}^{x} = \frac{(1-x)^2}{2} \cdot \frac{x^3}{2} \]Simplifying the result yields:\[ \frac{x^4(1-x)^2}{4} \]
05

Perform the Integration with Respect to x

Integrate the expression with respect to x:\[ \int_{0}^{1} \frac{x^4(1-x)^2}{4} \, dx = \frac{1}{4} \int_{0}^{1} x^4(1 - 2x + x^2) \, dx \]Expanding the terms:\[ \frac{1}{4} \left( \int_{0}^{1} x^4 \, dx - 2 \int_{0}^{1} x^5 \, dx + \int_{0}^{1} x^6 \, dx \right) \]Calculating each integral separately:\[ \frac{1}{4} \left( \left[ \frac{x^5}{5} \right]_{0}^{1} - 2 \left[ \frac{x^6}{6} \right]_{0}^{1} + \left[ \frac{x^7}{7} \right]_{0}^{1} \right) \]\[ = \frac{1}{4} \left( \frac{1}{5} - \frac{1}{3} + \frac{1}{7} \right) \]Simplifying the expression:\[ = \frac{1}{4} \cdot \frac{21 - 35 + 15}{105} \]\[ = \frac{1}{4} \cdot \frac{1}{35} = \frac{1}{140} \]
06

Conclusion

The evaluated triple integral results in a value of \( \frac{1}{140} \). This is the volume integral of the given function over the specified tetrahedron.

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Key Concepts

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

Tetrahedron
A tetrahedron is a three-dimensional geometric shape with four triangular faces, six edges, and four vertices. You can think of it as a pyramid with a triangular base. The problem at hand involves a particular tetrahedron defined by given vertices:
  • (0,0,0): The origin point, serving as one of the base points.
  • (1,0,0) and (1,1,0): These form the triangular base lying on the xy-plane.
  • (1,0,1): This is the apex point of the tetrahedron, above the xy-plane.
Each vertex of the tetrahedron marks a critical intersection of the planes forming its structure. To visualize, the base is a right triangle on the xy-plane. From this base, the peak of the tetrahedron rises up to (1,0,1), forming the other triangular faces. The complete solid is bounded by linear planes that define its edges. Understanding these boundaries is vital for the setting of subsequent integrals and performing volume calculations.
Volume Integral
A volume integral allows us to compute quantities like mass, charge, or in this case, a mathematical expression over a three-dimensional region鈥攊n our problem, a tetrahedron. The integral, \[ \iiint_{T} x y z \, dV \] is a triple integral: a sequential integral over a three-dimensional space, where each integration involves different variables (x, y, and z). For a thorough computation, this triple integral must be approached step-by-step:
  • Integrate with respect to z: Given the limits tied to the z-variable, we calculate first, considering z鈥檚 dependence on x, reducing the product function accordingly.
  • Integrate with respect to y: Once z is fixed and integrated, y is tackled, with results adjusting per the y constraints related to x.
  • Integrate with respect to x: Finally, solve for x, enveloping the entire calculation to output the integral's value over the tetrahedron.
By applying these integrations over the dictated limits, the integral uncovers a net value that reflects the total magnitude of the x, y, z configuration across the tetrahedron.
Limits of Integration
To solve any integral over a region, defining the correct limits of integration is crucial. It precisely states where the integration begins and ends on each axis. For our tetrahedron,
  • x-axis limits: Since the tetrahedron base on the xy-plane extends from (0,0) to (1,1), the limits are clearly 0 鈮� x 鈮� 1.
  • y-axis limits: Within those x limits, y ranges based on the line between (0,0) and (1,1), hence the constraint 0 鈮� y 鈮� x.
  • z-axis limits: With z determined by the height from the base, according to the plane from (0,0,0) to (1,0,1), it's bound by 0 鈮� z 鈮� 1 - x.
Utilizing these nested limits creates a flow from outermost to innermost integration: starting with x, moving to y, and concentrating finally upon z. Accurately built limits guarantee the integration processes effectively across this solid geometry, capturing the complete three-dimensional essence of the problem space.

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