Evaluate The Double Integral_d (x^2 + 2y) DA, D Is Bounded By Y = X, Y = X^3, X 0

Evaluate The Double Integral_d (x^2 + 2y) DA, D Is Bounded By Y = X, Y = X^3, X 0

In the realm of multivariable calculus, evaluating double integrals over specified regions is a fundamental skill that allows us to compute areas, volumes, and accumulated quantities. In this article, we will thoroughly analyze and compute the double integral ∫∫_D (x² + 2y) dA, where the region D is bounded by the curves y = x, y = x³, and the vertical line x = 0. This problem provides an excellent opportunity to explore region boundaries, set up integrals correctly, and employ substitution techniques to simplify calculations.

Understanding the Region D

Identifying the Boundaries

The region D is described by the following boundaries:

• The curve y = x (a straight line passing through the origin with slope 1)
• The curve y = x³ (a cubic curve passing through the origin)
• The vertical boundary x = 0 (the y-axis)

These boundaries enclose a region in the xy-plane that lies to the right of the y-axis, between the two curves y = x³ and y = x.

Visualizing the Region

To better understand the region D, consider the following points:

• At x = 0, both y = x and y = x³ pass through (0, 0).
• For x > 0, the line y = x is above the cubic y = x³ because for x > 0, x > x³ (since x³ < x for x in (0,1)).
• The curves intersect at the origin and at x = 1, where:

y = x = x³ => x = x³ => x³ - x = 0 => x(x² - 1) = 0 => x = 0 or x = 1

Thus, the two curves intersect at (0, 0) and (1, 1).

Region D Summary

• x ranges from 0 to 1.
• For each fixed x in [0, 1], y varies between y = x³ (lower boundary) and y = x (upper boundary).

Setting Up the Double Integral

Choosing the Order of Integration

Since the region is naturally bounded between y = x³ and y = x for x in [0, 1], it is straightforward to set up the double integral with respect to y first, and then x. The integral becomes:

∫x=0 to 1 ∫y=x³ to x (x² + 2y) dy dx

Alternatively, we could consider integrating with respect to x first, but given the region's description, the y-integration approach is cleaner.

Explicit Form of the Integral

The double integral over D is:

∫x=0^{1} ∫y=x³^{x} (x² + 2y) dy dx

This setup respects the boundaries and allows us to evaluate step-by-step.

Calculating the Inner Integral

Inner Integral with Respect to y

Compute:

I(x) = ∫y=x³^{x} (x² + 2y) dy

Since x is treated as a constant during integration, the integral becomes:

I(x) = ∫y=x³^{x} x² dy + ∫y=x³^{x} 2y dy

Calculating each part separately:

1. ∫ x² dy = x² y |_y=x³^{x} = x² (x - x³) = x² (x - x³)


2. ∫ 2y dy = y² |_y=x³^{x} = x² - (x³)² = x² - x⁶

Adding both parts:

I(x) = x² (x - x³) + x² - x⁶

Simplify:

I(x) = x²·x - x²·x³ + x² - x⁶ = x³ - x⁵ + x² - x⁶

Evaluating the Outer Integral

Integral with Respect to x

Now, the original integral reduces to:

∫0^{1} [x³ - x⁵ + x² - x⁶] dx

Break into separate integrals:

∫₀¹ x³ dx - ∫₀¹ x⁵ dx + ∫₀¹ x² dx - ∫₀¹ x⁶ dx

Calculate each:

1. ∫₀¹ x³ dx = (x⁴)/4 |₀^1 = 1/4


2. ∫₀¹ x⁵ dx = (x⁶)/6 |₀^1 = 1/6


3. ∫₀¹ x² dx = (x³)/3 |₀^1 = 1/3


4. ∫₀¹ x⁶ dx = (x⁷)/7 |₀^1 = 1/7

Putting it all together:

Total = 1/4 - 1/6 + 1/3 - 1/7

Combining the Fractions

Find the common denominator, which is 84 (LCM of 4, 6, 3, 7):

• 1/4 = 21/84
• 1/6 = 14/84
• 1/3 = 28/84
• 1/7 = 12/84

Compute:

Total = (21/84) - (14/84) + (28/84) - (12/84) = (21 - 14 + 28 - 12)/84 = (23)/84

Final value of the double integral:

∫∫D (x² + 2y) dA = 23/84

Conclusion and Summary

Key Takeaways

• The region D is bounded by y = x, y = x³, and x = 0, with x ranging from 0 to 1.
• Setting up the double integral with y as the inner variable is straightforward due to the natural bounds.
• The integral simplifies nicely by integrating term-by-term.
• The final value of the double integral over the specified region is 23/84.

Practical Implications

This process demonstrates how to approach double integrals over complex regions, emphasizing the importance of:

• Understanding the region's boundaries
• Choosing an appropriate order of integration
• Simplifying the integrand before integration
• Carefully calculating and combining fractional results

Additional Tips for Evaluating Double Integrals

• Sketch the Region: Always draw the region to visualize boundaries and intersections.
• Select the Integration Order: Choose the order that simplifies calculations based on the region's shape.
• Set Up Limits Carefully: Ensure the limits reflect the exact region boundaries.
• Handle the Integrand: Simplify the integrand if possible before integrating.
• Compute Step-by-Step: Break the integral into manageable parts, especially when dealing with polynomial expressions.

Final Thoughts

Evaluating double integrals over regions bounded by curves such as y = x and y = x³ provides valuable experience in multivariable calculus. By carefully analyzing the region, setting up the integral correctly, and methodically performing the calculations, students and practitioners can solve complex integrals with confidence. The specific problem discussed confirms that with proper setup and algebraic simplification, even seemingly complicated integrals can be computed efficiently, leading to precise results like the value of 23/84 for the integral in question.