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Chapter 15: Problem 36

Write five other iterated integrals that are equal to the given iterated integral. $$ \int_{0}^{1} \int_{0}^{x^{2}} \int_{0}^{y} f(x, y, z) d z d y d x $$

Short Answer

Expert verified
Five iterated integrals are found by changing the order of integration while respecting the limits.

Step by step solution

01

Identify the Given Integral Limits

The given iterated integral is \( \int_{0}^{1} \int_{0}^{x^{2}} \int_{0}^{y} f(x, y, z) \; dz \; dy \; dx \). Here, the integration is performed in the order of \( dz, dy, dx \) with limits:\[\begin{align*} z: & \quad 0 \leq z \leq y, \ y: & \quad 0 \leq y \leq x^2, \ x: & \quad 0 \leq x \leq 1. \end{align*}\]We need to write this integral in five other different orders while ensuring the region of integration is preserved.
02

Reverse the Order to dy, dz, dx

First, change the order to \( dy \; dz \; dx \). To do this, integrate \( y \) from \( 0 \) to \( x^2 \), \( z \) from \( 0 \) to \( y \), and \( x \) from \( 0 \) to \( 1 \). The iterated integral becomes:\[\int_{0}^{1} \int_{0}^{x^2} \int_{0}^{y} f(x, y, z) \; dy \; dz \; dx.\]
03

Change the Order to dx, dz, dy

For \( dx \; dz \; dy \), \( x \) depends on \( y \), \( z \) on \( x \), and \( y \) as the top level. Thus, \( x \) integrates from \( 0 \) to \( \sqrt{y} \), \( z \) from \( 0 \) to \( y \), and \( y \) from \( 0 \) to \( 1 \) (as \( y \leq 1 \)). The integral becomes:\[\int_{0}^{1} \int_{0}^{x^2} \int_{0}^{\sqrt{y}} f(x, y, z) \; dx \; dz \; dy.\]
04

Change the Order to dz, dx, dy

Change integration order to \( dz \; dx \; dy \). Integrate \( x \) and \( y \) similarly to maintain bounds. Thus, \( z \) is from \( 0 \) to \( y \), \( x \) from \( 0 \) to \( \sqrt{y} \), and \( y \) from \( 0 \) to \( 1 \). The integral becomes:\[\int_{0}^{1} \int_{0}^{x^2} \int_{0}^{y} f(x, y, z) \; dz \; dx \; dy.\]
05

Change the Order to dx, dy, dz

For order \( dx \; dy \; dz \), \( x \) depends on \( y \), \( y \) on \( z \), and finally \( z \) independent. So, \( x \) integrates from \( 0 \) to \( 1 \), \( y \) from \( z \) to \( \sqrt{1} \), and \( z \) from \( 0 \) to \( 1 \). The iterated integral becomes:\[\int_{0}^{1} \int_{0}^{1} \int_{z}^{1} f(x, y, z) \; dx \; dy \; dz.\]
06

Change the Order to dy, dx, dz

Finally, consider \( dy \; dx \; dz \), integrating \( y \) from \( 0 \) to \( \sqrt{z} \), \( x \) from \( \sqrt{y} \) to \( 1 \), and \( z \) from \( 0 \) to \( 1 \). Thus, \[\int_{0}^{1} \int_{z}^{1} \int_{0}^{\sqrt{z}} f(x, y, z) \; dx \; dy \; dz.\]

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Key Concepts

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

Integration Order
When dealing with iterated integrals, the **order of integration** refers to the sequence in which we perform integrations over multiple variables. In the given problem, integrals were initially solved in the order \(dz, dy, dx\). However, you can rearrange the order in various ways to get the same final result. The key is to change the integration limits accordingly, to preserve the region of integration.

Maintaining the correct integration bounds for each variable is crucial. Different orderings can simplify the calculation process, depending on the integrand and the region of integration. Common orders include:
  • \(dz, dy, dx\)
  • \(dy, dz, dx\)
  • \(dx, dz, dy\)
Each order must account for dependencies between variables to keep the process mathematically sound.
Triple Integration
**Triple integration** is a technique used to compute the integral of a function over a three-dimensional space. This method involves integrating with respect to three separate variables, usually \(x\), \(y\), and \(z\).

Each variable corresponds to a distinct dimension:
  • \(x\)-axis for width
  • \(y\)-axis for height
  • \(z\)-axis for depth
In this exercise, the function \(f(x, y, z)\) is integrated over a specific 3D region defined by the limits of integration. The complete process is comprehensive, accounting for changes across three dimensions, enabling the evaluation of volumes and areas under curves or surfaces in space.
Integration Bounds
The **integration bounds** specify the limits over which the function is integrated and are essential for accurately defining the region of integration. In a triple integral, these boundaries cover three separate dimensions, and they can be functions of one or two of the other variables.

In the original exercise:
  • \(z\) ranges from \(0\) to \(y\)
  • \(y\) ranges from \(0\) to \(x^2\)
  • \(x\) ranges from \(0\) to \(1\)
Changing the order of integration involves adjusting these limits to ensure the same region is covered. For example, if \(dz\) is the outermost integral instead of the innermost, the bounds for \(z\) would need to be adjusted accordingly based on how \(y\) and \(x\) are defined later.
Multiple Integrals
**Multiple integrals** refer to integrals that involve more than one variable and are used to find volumes under surfaces or more complex shapes. Each integral corresponds to a dimension, with double integrals handling two dimensions and triple integrals handling three, as seen in this exercise.

Solving multiple integrals requires careful consideration of:
  • Order of integration
  • Changing integration limits appropriately for different orders
  • Reevaluating limits if the order changes
Understanding multiple integrals provides the footing for advanced calculus topics and applications, such as flux in vector calculus, center of mass calculations, or evaluating probability distributions over multi-dimensional spaces.

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Most popular questions from this chapter

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