Problem 27 - Thomas Calculus

Question

In Exercises \(21-30,\) sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$ \int_{0}^{3 / 2} \int_{0}^{9-4 x^{2}} 16 x d y d x $$

Step-by-Step Solution

Answer

Reverse the integral to: \( \int_{0}^{9} \int_{0}^{\sqrt{\frac{9-y}{4}}} 16x \, dx \, dy \).

Step 1: Understand the Problem

The problem involves reversing the order of a double integral. Currently, the integration is with respect to \( y \) first, over the interval \( [0, 9 - 4x^2] \), then with respect to \( x \) over the interval \( [0, \frac{3}{2}] \). Our task is to sketch the region of integration and then rewrite the integral with reversals.

Step 2: Identify the Region of Integration

The limits for \( y \) depend on \( x \), ranging from 0 to \( 9 - 4x^2 \). For \( x \), the range is from 0 to \( \frac{3}{2} \). This implies the region is bounded by the curves \( y = 0 \) and \( y = 9 - 4x^2 \), and by \( x = 0 \) and \( x = \frac{3}{2} \). This describes a region where \( 0 \leq x \leq \frac{3}{2} \) and \( 0 \leq y \leq 9 - 4x^2 \).

Step 3: Express the Bounds in Terms of y

To reverse the order of integration, solve for \( x \) in terms of \( y \). The curve \( y = 9 - 4x^2 \) can be rewritten as \( x = \pm \sqrt{\frac{9 - y}{4}} \). Given \( x \ge 0 \), we choose the positive root: \( x = \sqrt{\frac{9 - y}{4}} \). The bounds for \( x \) are \( 0 \le x \le \sqrt{\frac{9 - y}{4}} \).

Step 4: Determine Y-Range

The value of \( y \) ranges from the minimum obtained when \( x = 0 \), resulting in \( y = 0 \), to the maximum when \( x = \frac{3}{2} \), resulting in \( y = 9 - 4(\frac{3}{2})^2 = 0 \). Hence, \( y \) ranges from 0 to 9.

Step 5: Write the Reversed Integral

With the bounds identified, the integral with reversed order becomes \( \int_{0}^{9} \int_{0}^{\sqrt{\frac{9-y}{4}}} 16x \, dx \, dy \). This accounts for integrating with respect to \( x \) first, then \( y \).

Step 6: Verification and Sketch

Ensure the region sketched matches the bounds. This involves confirming the parabolic shape of \( (y = 9 - 4x^2) \) within the bounds of \( x = 0 \) to \( x = 1.5 \). The reversed integral now fits a rising curve from \( x^2 = 0 \) to \( y = 9 \).

Key Concepts

Reversing Order of Integration

Reversing the order of integration in a double integral means swapping the order in which we integrate with respect to each variable. In our exercise, the original integral is \[ \int_{0}^{3/2} \int_{0}^{9-4x^2} 16x \, dy \, dx \]where we have the variable \( y \) integrated first, followed by the variable \( x \). To reverse this, we will integrate with respect to \( x \) first and then with respect to \( y \).
Reversing integration involves adjusting the limits of integration to match the new order, ensuring that they describe the same region of integration.
To find the new limits:

• Solve the equation \( y = 9 - 4x^2 \) for \( x \), giving \( x = \sqrt{\frac{9-y}{4}} \) for \( x \geq 0\).
• Determine the range for \( y \), which goes from 0 to the upper bound found by substituting the maximum \( x \) value into the original equation.

The equivalent reversed integral reflects these adjustments, becoming \[ \int_{0}^{9} \int_{0}^{\sqrt{\frac{9-y}{4}}} 16x \, dx \, dy \].

Region of Integration

The region of integration in a double integral identifies the area over which the integration is performed. For this particular problem, the original region is defined by the inequalities:

• \( 0 \leq x \leq \frac{3}{2} \)
• \( 0 \leq y \leq 9 - 4x^2 \)

This means the region is bounded:
- Horizontally by the line \( x = 0 \) and \( x = \frac{3}{2} \), and- Vertically by \( y = 0 \) (the x-axis) and the curve \( y = 9 - 4x^2 \).
When reversing the order of integration, we need to re-evaluate this region to determine new bounds for \( x \) in terms of \( y \). This yields a new description of the region:

• \( 0 \leq y \leq 9 \)
• Within this, \( 0 \leq x \leq \sqrt{\frac{9-y}{4}} \)

Integration Bounds

Integration bounds are the specific limits between which the integration takes place, specifying the start and end points of integration for each variable. They vary depending on the order of integration:
In the original integral:

• The bounds for \( y \) are \( 0 \leq y \leq 9 - 4x^2 \).
• The bounds for \( x \) are \( 0 \leq x \leq \frac{3}{2} \).

When reversed:

• The bounds for \( x \) become \( 0 \leq x \leq \sqrt{\frac{9-y}{4}} \).
• The bounds for \( y \) are altered to \( 0 \leq y \leq 9 \), covering the span of the parabola's effect.

These bounds must precisely reflect the contours and constraints of the integration region, ensuring that all parts of the region are included once and only once.

Sketching Regions

Sketching the region of integration is a crucial visual step that aids in understanding the area covered by the double integral. It involves drawing the curves or lines that define the integration boundaries and highlighting the enclosed area.
For our problem, sketching begins by plotting the equation \( y = 9 - 4x^2 \), which is a downward-facing parabola opening along the y-axis, intersecting the y-axis at \( y = 9 \).

• Draw the vertical line \( x = 0 \) and the horizontal line \( y = 0 \), which are the other boundaries.
• The parabola intersects the x-axis exactly at \( x = \pm \frac{3}{2} \), since setting \( y = 0 \) gives this result.

By visually representing this region, you can effectively comprehend the area over which the integration occurs, making reversing the order easier by focusing on how the area bounds change.