Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 33-38, write the given iterated integral as an iterated integral with the order of integration interchanged. Hint: Begin by sketching a region \(S\) and representing it in two ways, as in Example \(5 .\) \(\int_{-1}^{0} \int_{-\sqrt{y+1}}^{\sqrt{y+1}} f(x, y) d x d y\)

Step-by-Step Solution

Answer

The interchanged integral is \( \int_{-1}^{1} \int_{-1}^{x^2 - 1} f(x, y) \, dy \, dx \).

Step 1: Sketch the Region of Integration

First, we need to understand the limits of integration by sketching the region over which the integration is carried out. The inner integral \( \int_{-\sqrt{y+1}}^{\sqrt{y+1}} \) indicates that for each \( y \), \( x \) ranges from \(-\sqrt{y+1}\) to \(\sqrt{y+1}\). The outer integral \( \int_{-1}^{0} \) indicates that \( y \) ranges from \(-1\) to \(0\). This suggests a region bounded by the parabola \( x = \pm \sqrt{y+1} \) for \( y \in [-1, 0] \).

Step 2: Express the Region with Reversed Limits

Once the region of integration \( S \) is understood, interchange the integral order by expressing \( x \) as the outer limit. The parabola \( x = \pm \sqrt{y+1} \) can be restated as \( y = x^2 - 1 \). The range of \( x \) in the original order is from \(-1\) to \(1\), calculated by substituting\( y = 0 \) (the upper limit). Thus, integrate \( y \) next from \( -1 \) to the function \( x^2 - 1 \).

Step 3: Write the Reordered Integral

Using the limits determined in the previous step, the iterated integral with reversed order is \( \int_{-1}^{1} \int_{-1}^{x^2 - 1} f(x, y) \, dy \, dx \). Here, \( x \) varies from \(-1\) to \(1\), and for each \( x \), \( y \) varies from \(-1\) to \( x^2 - 1 \). This setup mirrors the geometric region initially described, but with the order of integration swapped.

Key Concepts

Order of Integration

In the world of iterated integrals, the order of integration refers to the sequence in which you integrate with respect to the variables involved. Typically, iterated integrals are written as double or triple integrals where each integral is with respect to a different variable. By changing the order of integration, you might be able to simplify the integral, making calculations easier or even making it possible to integrate in the first place.
In our specific exercise, the original integral is \[\int_{-1}^{0} \int_{-\sqrt{y+1}}^{\sqrt{y+1}} f(x, y) dx \, dy\].

This double integral first integrates with respect to \(x\) (across the lines of the inner limits \(- \sqrt{y+1}\) to \(\sqrt{y+1}\)), and then with respect to \(y\) (determined by the outer limits \(-1\) to \(0\)). By swapping the order, the integration sequence changes. Now, the integral becomes:
\[\int_{-1}^{1} \int_{-1}^{x^2 - 1} f(x, y) \, dy \, dx\].
This reordering starts integrating with respect to \(y\) first, from the lower bound \(-1\) to \(x^2 - 1\), followed by \(x\) from \(-1\) to \(1\). This new order can turn a difficult integral into a more approachable one.

Sketching Regions

Visualizing the region of integration is an indispensable step in solving for iterated integrals, especially when changing the order of integration. By sketching, you bring the abstract concept of limits into a concrete visual interpretation. This helps determine whether or not the regions and boundaries are correctly understood.
In this exercise, we are dealing with a region defined by the inequalities:

• \(-1 \leq y \leq 0\)
• \(-\sqrt{y+1} \leq x \leq \sqrt{y+1}\)

The equation \(x = \pm \sqrt{y+1}\) forms a parabola opening sideways, intersecting the \(y\)-axis between \(-1\) and \(0\).
Drawing this region helps visualize how to reframe the integration limits when swapping the order of integration. Now, after swapping, the bounds are viewed in terms of \(x\):

• \(-1 \leq x \leq 1\)
• \(-1 \leq y \leq x^2 - 1\)

This picture supports setting up your new integral, ensuring that every point in your originally sketched region is accurately represented.

Integration Limits

Determining the correct limits of integration is crucial for evaluating iterated integrals. Limits define the boundary within which the integration occurs, dictating where your function is evaluated.
In our integral setup, the original limits for \(x\) were given by \(-\sqrt{y+1}\) to \(\sqrt{y+1}\), and for \(y\), the limits were \(-1\) to \(0\). These limits are based on the shape and constraints of the region bounded by the parabola \(x = \pm \sqrt{y+1}\).
Upon reversing integration order, the new system of bounds reflects a different structure:

• For \(x\), it spans from \(-1\) to \(1\), indicating the full width of the double-petal of the parabola.
• Within these \(x\) boundaries, \(y\) ranges from \(-1\) to \(x^2 - 1\), illustrating the top part of the resulting parabolic slice.

Each new limit is essential for covering the same area with a different starting point, ensuring accuracy in evaluation.

Parabola in Integration

Parabolas often appear in regions of integration due to their smooth, symmetrical shapes, which can effectively form curved boundaries helpful in calculus.
Here, the parabola defined by \(x = \pm \sqrt{y+1}\) presents a common scenario where integration takes place within a curvilinear domain.
This parabola forms a sideways opening, creating a bounded region within \(y = -1\) and \(y = 0\) on the coordinate plane. When considering changing the order of integration, recognizing the parabola's shape lets us redefine the limits in terms of this geometric feature.
Re-expressed, the region becomes bounded on top by the symmetric parabolic form \(y = x^2 - 1\), a transformation crucial when swapping limits to become:
\[-1 \leq x \leq 1\]

• The previously vertical symmetry easily converts into horizontal constraints, delineating the integration domain anew.
• This shape is pivotal in restructuring limits based on symmetrical and quadratic order, working hand-in-hand with specified boundaries.

Understanding this parabolic influence grants greater ease when addressing integration within its confines.