/*! 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 9 Evaluate the iterated integrals.... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 9

Evaluate the iterated integrals. \(\int_{-2}^{4} \int_{x-1}^{x+1} \int_{0}^{\sqrt{2 y / x}} 3 x y z d z d y d x\)

Short Answer

Expert verified
The value of the iterated integral is 156.

Step by step solution

01

Evaluate the Innermost Integral

First, integrate with respect to \(z\). The innermost integral is \(\int_{0}^{\sqrt{2y/x}} 3xyz\, dz\).\ Since 3xy is a constant with respect to \(z\), we integrate \(z\) to get \(\frac{3xyz^2}{2}\) evaluated from \(0\) to \(\sqrt{2y/x}\).\[ \int_{0}^{\sqrt{2y/x}} 3xyz\, dz = \left[ \frac{3xyz^2}{2} \right]_{0}^{\sqrt{2y/x}} = \frac{3xy(2y/x)}{2} = 3y^2. \]
02

Evaluate the Middle Integral

Next, integrate the result from Step 1 with respect to \(y\).\[ \int_{x-1}^{x+1} 3y^2\, dy. \] The antiderivative of \(3y^2\) is \(y^3\), so we evaluate it between \(x-1\) and \(x+1\):\[ \left[ y^3 \right]_{x-1}^{x+1} = (x+1)^3 - (x-1)^3. \]
03

Simplify the Result of the Middle Integral

Simplify the expression \((x+1)^3 - (x-1)^3\): \[ (x+1)^3 = x^3 + 3x^2 + 3x + 1,\] \[ (x-1)^3 = x^3 - 3x^2 + 3x - 1. \] Thus, \( (x+1)^3 - (x-1)^3 = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1) = 6x^2 + 2.\)
04

Evaluate the Outermost Integral

Finally, integrate the result \(6x^2 + 2\) with respect to \(x\) from \(-2\) to \(4\):\[ \int_{-2}^{4} (6x^2 + 2)\, dx. \] Compute the antiderivative: \(\int 6x^2\, dx = 2x^3\) and \( \int 2\, dx = 2x\). Evaluate the antiderivative from \(-2\) to \(4\):\[ \left[ 2x^3 + 2x \right]_{-2}^{4} = \left(2(4)^3 + 2(4)\right) - \left(2(-2)^3 + 2(-2)\right).\]
05

Calculate the Final Result

Compute the above expression:\[ \left(2(4)^3 + 2(4)\right) = 2 \times 64 + 8 = 136. \] \[ \left(2(-2)^3 + 2(-2)\right) = 2 \times (-8) - 4 = -20. \] Thus, the final result is \(136 - (-20) = 136 + 20 = 156.\)

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.

Multiple Integration
Multiple integration extends the concept of single-variable integration to functions of multiple variables. When dealing with real-world problems, these functions usually depend on two or more variables.
For example, surfaces in a 3D space. Multiple integration helps compute volumes, areas, and other properties that involve multiple dimensions.
Iterated integrals are evaluations of these multiple integrations, and they involve performing the integration process in a step-by-step manner.
  • Typically, we first deal with the innermost integral, then progress outward.
  • Each step requires integration of the function with respect to a different variable, while treating other variables as constants.
In the given problem, the triple integral involves integration with respect to three variables: \(z\), \(y\), and \(x\), in that order.
Calculus
Calculus is a fundamental branch of mathematics that is used to understand changes between values related through functions. In the context of our problem, it provides the rules and notations that let us perform multiple integrals.
Calculus consists of two main branches:
  • Differential calculus, which deals with the rate of change (derivatives).
  • Integral calculus, which involves finding the total size or value, such as areas under curves (integrals).
In our problem, integral calculus is more relevant, as it allows us to find the volume under a surface described by the function \(3xyz\).
The iterated integral lets us break down this problem into sequential steps, each focusing on one variable at a time.
Definite Integral
A definite integral gives the accumulated sum of areas under or above a curve, bound by a specific range.
It gives a numerical value instead of a function, which makes it distinct from the indefinite integral, which describes a family of functions.
In our exercise, each step of integration with respect to \(z\), \(y\), and \(x\) is a definite integral with specific limits.
  • The innermost integral limits for \(z\) are from \(0\) to \(\sqrt{2y/x}\).
  • The middle integral limits for \(y\) range from \(x-1\) to \(x+1\).
  • The outermost integral limits for \(x\) are from \(-2\) to \(4\).
Definite integrals provide the final value of 156 when we combine all steps.
Integration Techniques
Integration can sometimes be challenging, so multiple techniques have been developed to help in solving integrals.
These techniques include but are not limited to substitution, integration by parts, and partial fractions.
In our problem, the integration technique involved the sequential approach of evaluating iterated integrals.
  • Each step simplifies to a basic integral, often easy to solve using well-known formulas.
  • The calculation involves consistently plugging in the limits to evaluate the integral at each step.
Simplification skills are vital as seen in the steps where we simplified polynomial expressions.
This approach effectively breaks down complex integral problems into manageable parts.

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

Suppose that the random variables \((X, Y)\) have joint PDF $$ f(x, y)=\left\\{\begin{array}{ll} k y, & \text { if } 0 \leq x \leq 12 ; 0 \leq y \leq x \\ 0, & \text { otherwise } \end{array}\right. $$ Find each of the following: (a) \(\underline{k}\) (b) \(P(Y>4)\) (c) \(E(X)\)

Let \(S_{1}\) and \(S_{2}\) be disjoint laminas in the \(x y\) -plane of mass \(m_{1}\) and \(m_{2}\) with centers of mass \(\left(\bar{x}_{1}, \bar{y}_{1}\right)\) and \(\left(\bar{x}_{2}, \bar{y}_{2}\right) .\) Show that the center of mass \((\bar{x}, \bar{y})\) of the combined lamina \(S_{1} \cup S_{2}\) satisfies $$ \bar{x}=\bar{x}_{1} \frac{m_{1}}{m_{1}+m_{2}}+\bar{x}_{2} \frac{m_{2}}{m_{1}+m_{2}} $$ with a similar formula for \(\bar{y}\). Conclude that in finding \((\bar{x}, \bar{y})\) the two laminas can be treated as if they were point masses at \(\left(\bar{x}_{1}, \bar{y}_{1}\right)\) and \(\left(\bar{x}_{2}, \bar{y}_{2}\right)\).

Find the moments of inertia \(I_{x}, I_{y}\), and \(I_{z}\) for the lamina bounded by the given curves and with the indicated density \(\delta .\) \(y=\sqrt{x}, x=9, y=0 ; \delta(x, y)=x+y\)

Find the area of the indicated surface. Make a sketch in each case. The surface of the solid that is the intersection of the two solid cylinders \(x^{2}+z^{2} \leq a^{2}\) and \(x^{2}+y^{2} \leq a^{2} .\) Hint: You may need the integration formula \(\int(1+\sin \theta)^{-1} d \theta=\) \(-\tan [(\pi-2 \theta) / 4]+C\).

Use spherical coordinates to find the indicated quantity. Mass of a solid inside a sphere of radius \(2 a\) and outside a circular cylinder of radius \(a\) whose axis is a diameter of the sphere, if the density is proportional to the square of the distance from the center of the sphere
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Mechanics 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.