Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Write the given iterated integral as an iterated integral with the indicated order of integration. \(\int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} \int_{0}^{\sqrt{1-y^{2}-z^{2}}} f(x, y, z) d x d z d y ; d z d y d x\)

Step-by-Step Solution

Answer

Rewritten integral: \(\int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}} f(x, y, z) dz dy dx\).

Step 1: Understand Current Integral Order

The given iterated integral is ordered as follows: first integrate over \(x\), then \(z\), and finally \(y\). This is denoted by \(dx\, dz\, dy\).

Step 2: Identify New Desired Order

The problem asks us to rewrite the integral so the new order of integration is \(dz\, dy\, dx\). This means that we want to integrate with respect to \(z\) first, then \(y\), and finally \(x\).

Step 3: Express the Region of Integration

The original integral integrates over the region where \(0 \leq x \leq \sqrt{1-y^2-z^2}\), \(0 \leq z \leq \sqrt{1-y^2}\), and \(0 \leq y \leq 1\). This describes a portion of a sphere.

Step 4: Change Integral Order

To find the bounds for this new order, we consider: - For \(z\): it varies from \(0\) to the surface defined by \(x^2+y^2+z^2 \leq 1\), meaning \(z\) goes from \(-\sqrt{1-x^2-y^2}\) to \(+\sqrt{1-x^2-y^2}\).- For \(y\): it varies such that \(0 \leq y \leq \sqrt{1-x^2}\).- For \(x\): it retains its original bounds, from \(0\) to \(1\).

Step 5: Write the New Integral

Combining the newly found limits, the iterated integral can be rewritten as:\[\int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}} f(x, y, z) dz dy dx\]

Key Concepts

Understanding Order of Integration

When dealing with iterated integrals, the **order of integration** is crucial. It tells us the sequence in which we compute the integrals. Usually, you have multiple nested integrals, and each one corresponds to integrating a different variable one after the other. Changing the order of integration is not just a matter of rearranging terms. It often involves reevaluating bounds based on how each variable depends on one another.

In the original exercise, the order was given as \(dx\, dz\, dy\), meaning \(x\) is integrated first, then \(z\), and finally \(y\). The task was to switch to \(dz\, dy\, dx\). When you change the order, the integral's **bounds** must be recalculated to reflect the new order, ensuring the same region in three-dimensional space is still being integrated over. This can be complex because it involves how these variables interact spatially.

Triple Integrals Demystified

**Triple integrals** take integration into three dimensions. This is an essential tool in calculus, especially when analyzing regions or volumes in 3D space. Each integral corresponds to one of the three dimensions:

• First integral: represents the first variable
• Second integral: represents the second variable
• Third integral: the last variable

These integrals are used to calculate volumes, masses, and other quantities where a region in three-dimensional space is involved.

The exercise's integral moves through a 3D region, first integrating over \(x\), then \(z\), and then \(y\). By switching to the new order \(dz\, dy\, dx\), you are still covering the same region but recalculating how you pass through the space. This does not change the integral's value, provided the bounds match the same physical region.

Exploring the Region of Integration

The **region of integration** in a multiple integral defines where you're calculating the function's value. It outlines the limits within which each variable is permitted to move. For triple integrals, this region becomes a volume in three-dimensional space.

In the original problem, this region is bounded by the surfaces and equations given: \\(0 \leq x \leq \sqrt{1-y^2-z^2}\), \(0 \leq z \leq \sqrt{1-y^2}\), and \(0 \leq y \leq 1\). This particular region is part of a **sphere**.

Because the integral bounds form regions constrained by squared terms, they describe portions of a solid object, here a sphere. Understanding the geometric shape these bounds describe is essential for changing orders, as the bounds translate into coordinates and need recalibration for different sequences.

Applying Spherical Coordinates

While spherical coordinates are not directly used in this integral, knowing about them is helpful. **Spherical coordinates** are a system of three numbers that specify a point's location based on its distance from a reference point and two angles. They are particularly advantageous in problems involving spheres or spherical regions.

They consist of:

• \(\rho\): the radius or distance from the origin
• \(\theta\): the angle from the positive \(x\)-axis in the \(xy\)-plane
• \(\phi\): the angle from the positive \(z\)-axis

These are especially useful for converting complex volume integrals into simpler forms. Although the exercise refrains from spherical coordinates directly, recognizing the spherical nature of the region (due to \(x^2 + y^2 + z^2\)) can make solving complex integrals easier by depicting the volume in familiar, calculable shapes.