Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Write the given iterated integral as an iterated integral with the indicated order of integration. \(\int_{0}^{2} \int_{0}^{9-x^{2}} \int_{0}^{2-x} f(x, y, z) d z d y d x ; d y d x d z\)

Step-by-Step Solution

Answer

Rewritten integral: \( \int_{0}^{2} \int_{0}^{2-z} \int_{0}^{9-(2-z)^{2}} f(x, y, z) \ dy \ dx \ dz \).

Step 1: Understand Initial Order

The original integral is given as \( \int_{0}^{2} \int_{0}^{9-x^{2}} \int_{0}^{2-x} f(x, y, z) \ d z \ d y \ d x \), with the order of integration as \( dz \ dy \ dx \). This means \( z \) is integrated from \( 0 \) to \( 2-x \), then \( y \) from \( 0 \) to \( 9-x^{2} \), and finally \( x \) from \( 0 \) to \( 2 \).

Step 2: Determine the New Order

We want to change the order to \( dy \ dx \ dz \). This requires us to rewrite the integrals in this new order while maintaining the limits of integration that correctly describe the region of integration.

Step 3: Identify the Bounds for z

In the original setup, \( z \) is limited by \( 0 \leq z \leq 2-x \). For the new order, \( z \) will vary from the minimum possible value, \( 0 \), to its maximum, \( 2 \). Since \( z \) is innermost in the new order, it becomes the outermost, so \( 0\leq z\leq 2 \).

Step 4: Express x in terms of z

From \( z = 2-x \), we solve for \( x \) to get \( x = 2-z \). The limits for \( x \) based on this expression and the original bounds ensure that \( x \) starts from \( 0 \) and goes up to \( 2-z \).

Step 5: Establish the Bounds for y

Originally, \( y \) was between \( 0 \) and \( 9 - x^2 \). Under the new order, \( x \) dictates \( y \)'s bounds. Thus, for any \( x \), \( y \) ranges from \( 0 \) to \( 9 - x^2 \), and since \( x \) depends on \( z \) with the expression \( x = 2-z \), these limits are valid.

Step 6: Write the New Integral

Combining the limits found and the expression for \( x \), the integrals become \( \int_{0}^{2} \int_{0}^{2-z} \int_{0}^{9-(2-z)^{2}} f(x, y, z) \ dy \ dx \ dz \), with appropriate substitutions.

Key Concepts

Order of Integration

The order of integration refers to the sequence or arrangement in which multiple integrals are solved within an iterated integral. Iterated integrals involve resolving one integral at a time, moving gradually through the nested structure.
Initially, our problem involves the integral order of \(dz \ dy \ dx\), which dictates that we integrate with respect to \(z\) first, \(y\) second, and \(x\) last. The challenge here is to reorder this integral to \(dy \ dx \ dz\).

• Reordering can sometimes simplify the integration process.
• Each order change could directly affect the integration bounds for each variable.

Changing the order requires careful attention to the limits of integration, as a misstep could lead to incorrect solutions.

Integration Bounds

Integration bounds define the limits through which each variable ranges during the integration process. In multivariable calculus, these bounds can be functions, constants, or a mix.
In our original exercise, the bounds were:

• For \(z\): from 0 to \(2 - x\).
• For \(y\): from 0 to \(9 - x^2\).
• For \(x\): from 0 to 2.

When changing the order to \(dy \ dx \ dz\), the bounds must adjust to reflect the regions accurately:

• \(z\) will now range from 0 to 2, acting as the outermost integral.
• \(x\) limits depend on \(z\), from 0 to \(2-z\).
• \(y\), being the innermost, will range from 0 to \(9 - (2-z)^2\).

Each trajectory a variable takes is crucial for accurately defining the space over which we're integrating.

Change of Variables

Changing variables in integrals is a powerful method that helps simplify complex integrals and makes it possible to change the order of integration. The step where we solve \(z = 2-x\) to find \(x = 2-z\) is a classic case of changing variables.
This transformation leads to the adjustment of integration limits and the expression for variables within the integral.

• By solving for \(x\) in terms of \(z\), we can adjust \(x\)'s limits correctly based on \(z\).
• This change ensures that our volume of integration remains consistent, even as we alter the sequence of integration.

Such changes are essential in iterated integrals to achieve accurate and solvable forms that align with the new order of integration.

Calculus

Calculus is the mathematical study of change, and it forms the backbone of solving integrals, especially in multiple dimensions. When dealing with iterated integrals, calculus allows us to tackle complex problems involving three-dimensional spaces. Using calculus:

• We determine how variables influence one another within an integral.
• We apply integration techniques sequentially to derive solutions.
• We leverage fundamental theorems to simplify processes.

In this exercise, calculus enables us to transition from solving a simple integral to understanding the geometric volume described in the iterated integral. Understanding these principles helps to appreciate the full aspect of integration and its applications in solving real-world problems.