/*! 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 12 Evaluate the integrals in Exerci... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 12

Evaluate the integrals in Exercises \(7-20\). $$\int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1}(x+y+z) d y d x d z$$

Short Answer

Expert verified
The integral evaluates to 0.

Step by step solution

01

Understand the Problem

We are asked to evaluate the triple integral \( \int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} (x+y+z)\, dy\, dx\, dz \). This means we need to integrate the function \( x+y+z \) over a cube defined by \( -1 \leq x, y, z \leq 1 \).
02

Evaluate the Inner Integral with Respect to y

We first integrate the function \( x+y+z \) with respect to \( y \) from \( -1 \) to \( 1 \): \[ \int_{-1}^{1} (x + y + z) \, dy = \left[ xy + \frac{y^2}{2} + yz \right]_{-1}^{1} = \left( x(1) + \frac{1^2}{2} + z(1) \right) - \left( x(-1) + \frac{(-1)^2}{2} + z(-1) \right) \]Simplify further: \[ = (x + \frac{1}{2} + z) - (-x + \frac{1}{2} - z) = 2x + 2z \]
03

Evaluate the Middle Integral with Respect to x

Next, integrate \( 2x + 2z \) with respect to \( x \) from \( -1 \) to \( 1 \):\[ \int_{-1}^{1} (2x + 2z) \, dx = \left[ x^2 + 2zx \right]_{-1}^{1} = \left( 1^2 + 2z(1) \right) - \left( (-1)^2 + 2z(-1) \right) \]Simplify:\[ = (1 + 2z) - (1 - 2z) = 4z \]
04

Evaluate the Outer Integral with Respect to z

Finally, integrate \( 4z \) with respect to \( z \) from \( -1 \) to \( 1 \):\[ \int_{-1}^{1} 4z \, dz = \left[ 2z^2 \right]_{-1}^{1} = 2(1^2) - 2((-1)^2) = 2 - 2 = 0 \]
05

Conclusion

The value of the triple integral \( \int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} (x+y+z) \, dy \, dx \, dz \) over the specified limits is 0.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Iterated Integrals
Iterated integrals are a method used to evaluate multiple integrals by performing integration one variable at a time in a nested sequence. In the given problem, we have a triple integral which involves integrating with respect to three variables:\( x, y, \, \text{and} \, z \). This means we perform an integral with respect to one variable while treating the others as constants, and repeat the process for each variable.
Here's the order of integration:
  • First, integrate with respect to \( y \)
  • Second, integrate the resulting function with respect to \( x \)
  • Finally, integrate the resulting expression with respect to \( z \)
This systematic approach allows us to simplify a complex integral by breaking it into manageable steps. Understanding the order in iterated integrals is essential to accurately solve problems involving more than one variable.
Integration Over a Cube
When integrating over a cube, it's important to consider the region of integration and the limits associated with each variable. The cube referred to in this exercise has boundaries defined by the limits \(-1 \leq x, y, z \leq 1\). This means our domain of integration is a perfect cube centered at the origin with side length of 2 units.
In practical terms, we are integrating the function \( x+y+z \) over all points within this cube. This provides a well-defined and bounded three-dimensional space, simplifying the evaluation process since all variables share the same symmetric limits.
Understanding the geometry and limits aids in effectively setting up and solving the triple integral.
Evaluation of Definite Integrals
Evaluating definite integrals involves calculating the "accumulated" value of a function over an interval, providing a clear result rather than an indefinite answer with constants.
For each step in our iterated integration:
  • After integrating with respect to \( y \): The result is \( 2x + 2z \).
  • We then integrate with respect to \( x \): This yields \( 4z \).
  • Finally, integrating with respect to \( z \) results in the ultimate value of \( 0 \).
In each step, we substitute the upper and lower limits for the variable of integration, applying the Fundamental Theorem of Calculus. This method provides a definitive solution, allowing you to calculate the exact value to be evaluated across the specified range.
Symmetric Limits
Symmetric limits in integration refer to limits that are equidistant from a central value, often simplifying calculations due to symmetry. In this problem, each variable \( x, y, \text{and } z\) ranges from \(-1\) to \(1\).
Such symmetry offers benefits, including:
  • Identifying even or odd functions. Here, \( x+y+z \) is odd over the symmetric limits. This symmetry can sometimes lead integrals, like in this case, to become zero.
  • Reducing computational complexity by providing consistent bounds for each variable.
  • Highlighting the central point (\(0\) in this case), allowing for simplifications in multivariable calculus when integrating functions evenly distributed about this point.
Recognizing and leveraging symmetric limits speeds up and simplifies the integration process.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An infinite half-cylinder Let \(D\) be the interior of the infinite right circular half-cylinder of radius 1 with its single-end face suspended 1 unit above the origin and its axis the ray from \((0,0,1)\) to \(\infty .\) Use cylindrical coordinates to evaluate $$ \iiint_{D} z\left(r^{2}+z^{2}\right)^{-5 / 2} d V $$

Cylinder and paraboloids Find the volume of the region bounded below by the paraboloid \(z=x^{2}+y^{2},\) laterally by the cylinder \(x^{2}+y^{2}=1,\) and above by the paraboloid \(z=\) \(x^{2}+y^{2}+1 .\)

a. Suppose that a solid right circular cone \(C\) of base radius \(a\) and altitude \(h\) is constructed on the circular base of a solid hemisphere \(S\) of radius \(a\) so that the union of the two solids resembles an ice cream cone. The centroid of a solid cone lies one-fourth of the way from the base toward the vertex. The centroid of a solid hemisphere lies three-eighths of the way from the base to the top. What relation must hold between \(h\) and \(a\) to place the centroid of \(C \cup S\) in the common base of the two solids?

In Exercises 13 and \(14,\) find a. the mass of the solid. b. the center of mass. A solid region in the first octant is bounded by the coordinate planes and the plane \(x+y+z=2 .\) The density of the solid is \(\delta(x, y, z)=2 x .\)

Reducing a double to a single integral By changing the order of integration, show that the following double integral can be reduced to a single integral: $$ \int_{0}^{x} \int_{0}^{u} e^{m(x-t)} f(t) d t d u=\int_{0}^{x}(x-t) e^{m(x-t)} f(t) d t $$ Similarly, it can be shown that $$ \int_{0}^{x} \int_{0}^{v} \int_{0}^{u} e^{m(x-t)} f(t) d u d v=\int_{0}^{x} \frac{(x-t)^{2}}{2} e^{m(x-t)} f(t) d t $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.