Question

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

Short answer

\( \int_{0}^{9} \int_{0}^{\sqrt{9-y}} \int_{0}^{2-x} f(x, y, z) \; dz \; dx \; dy \)

Step-by-step solution

Understand the Given Integral

The initial given integral is set up to integrate first with respect to \( z \), then \( y \), and finally \( x \). Thus, the limits are \( z: 0 \text{ to } 2-x \), \( y: 0 \text{ to } 9-x^2 \), and \( x: 0 \text{ to } 2 \).

Reordering the Integration

We need to rewrite the integral to integrate first with respect to \( z \), then \( x \), and finally \( y \), in the order \( dz \; dx \; dy \).

Determine the New Limits for \( z \)

The integral of \( z \) has an upper limit of \( 2-x \). Since \( x \) is also a variable limit, \( z \) remains between \( 0 \) and \( 2-x \).

Determine the New Limits for \( x \)

Since \( x \) depends on \( y \) and \( y \) has limits \( 0 \text{ to } 9-x^2 \), solve the inequality \( 0 \leq y \leq 9-x^2 \) for \( x \). Rearranging gives \( x^2 \leq 9-y \) implying \( x \) can range from \( 0 \) to \( \sqrt{9-y} \).

Determine the Limits for \( y \)

\( y \) has fixed limits based on the original integral, from \( 0 \) to \( 9 \text{ since } 0 \leq y \leq 9-x^2 \) across all permissible \( x \).

Write the New Integral

Given the new order and determined limits, the integral can now be written as: \[ \int_{0}^{9} \int_{0}^{\sqrt{9-y}} \int_{0}^{2-x} f(x, y, z) \; dz \; dx \; dy \] This matches the specified order \( dz \; dx \; dy \).

Key concepts

Order of Integration

In the world of calculus, the *order of integration* refers to the sequence in which multiple integrals are evaluated. When dealing with iterated integrals, especially those involving more than one variable, the order of integration can be crucial for solving the problem correctly. Changing the order can sometimes simplify integration or make it possible when it seems complex.

In the given problem, we start with the order \(dz\, dy\, dx\), meaning we first integrate with respect to \(z\), followed by \(y\), and lastly \(x\). The task is to change it to \(dz\, dx\, dy\). This requires adjusting the limits of integration accordingly since each variable is closely related to the others.

Understanding the original order helps identify how each variable is bounded by the others. As you shift the order, reassessing these bounds ensures the entire region of integration is accurately captured. This might seem tricky at first, but it's often a matter of recognizing the relationships between the variables and adjusting limits without changing the region you examine.

Triple Integrals

A *triple integral* is an integral extending over a three-dimensional space, evaluating a function of three variables. You might think of it as stacking layers of double integrals, as each layer adds another dimension. In our task, the integral extends over a region in \( \mathbb{R}^3 \), involving variables \( x, y, \) and \( z \).

Triple integrals are extremely useful for calculating volumes, mass, and other properties of 3D regions, where each point in the region contributes to the final integral value. When setting up a triple integral, you define bounding surfaces and limits for three separate integrations. The challenge lies in carefully determining these bounds, particularly when variables depend on each other.

In this context, the given integral simplifies the evaluation by first considering the smallest nested dimension \(z\), which affects \(x\), and so on, forming a layered approach. Thoughtful selection of integration order and boundaries ensures accurate evaluation of the entire space being examined.

Variable Limits of Integration

*Variable limits of integration* occur when the bounds of integration for one variable depend on the limits or values of another variable. This dynamic feature often arises in multi-variable calculus problems, where regions are described by curves or surfaces.

In the original integral, the variable limit appears as the bound for \(z\), which is \(2-x\). The limit of \(x\) depends on \(y\) as \(x\) ranges from \(0\) to \(\sqrt{9-y}\), while \(y\) has fixed limits from \(0\) to \(9\). As you adjust the order, these limits must also be recalculated to ensure the same region is covered.

Solving these types of integrals often involves geometric reasoning or substituting in one variable for another in the expression of limits – a bit like solving a puzzle. The ultimate goal is to break down a complex region into integrable parts. Carefully translating and reversing these limits can seem challenging but becomes clearer with practice and visualization of the region in question.