Problem 17 - Advanced Engineering Mathematics

Question

Evaluate the double integral over the region \(R\) that is bounded by the graphs of the given equations. Choose the most convenient order of integration. $$ \iint_{R} 2 x y d A ; y=x^{3}, y=8, x=0 $$

Step-by-Step Solution

Answer

The integral evaluates to 96.

Step 1: Understand the Region of Integration

Region \( R \) is bounded by the curves \( y = x^3 \), \( y = 8 \), and \( x = 0 \). First, determine the limits for \( x \) by finding the intersection of \( y = x^3 \) and \( y = 8 \), yielding \( x = 2 \). The limits for \( y \) are from \( 0 \) to \( 8 \), where \( x \) will vary from the limits suggested by the functions \( x = 0 \) to \( x = y^{1/3} \).

Step 2: Setting Up the Integral

Given that the region sets \( y = x^3 \) to \( y = 8 \) vertically, we choose to integrate with respect to \( y \) first. Therefore, the integral is set up as:\[\int_{0}^{2} \int_{x^3}^{8} 2xy\, dy\, dx.\]

Step 3: Evaluate the Inner Integral

First, evaluate the inner integral with respect to \( y \):\[\int_{x^3}^{8} 2xy\, dy = 2x \left[ \frac{1}{2} y^2 \right]_{x^3}^{8} = 2x \left( \frac{1}{2}(8)^2 - \frac{1}{2}(x^3)^2 \right) = 2x \left( 32 - \frac{x^6}{2} \right).\]Simplifying, this equals \( 64x - x^7 \).

Step 4: Evaluate the Outer Integral

Evaluate the outer integral with respect to \( x \):\[\int_{0}^{2} (64x - x^7)\, dx = \left[ 32x^2 - \frac{1}{8}x^8 \right]_{0}^{2} = \left( 32(2)^2 - \frac{1}{8}(2)^8 \right).\]Calculate this: \( 32 \times 4 - \frac{1}{8} \times 256 = 128 - 32 = 96 \).

Step 5: Conclusion

The value of the double integral \( \iint_{R} 2xy\, dA \) over the specified region is \( 96 \).

Key Concepts

Region of Integration

The region of integration, often referred to as the domain over which the integration is carried out, is a crucial concept in double integrals. In our problem, the region \( R \) is defined by several boundaries:

• The curve \( y = x^3 \)
• A horizontal line \( y = 8 \)
• The vertical line \( x = 0 \)

To visualize this region, one can plot these boundary lines on the xy-plane. This creates a window from \( x = 0 \) to \( x = 2 \) since \( y = x^3 \) intersects \( y = 8 \) at \( x = 2 \). Correspondingly, \( y \) varies from the curve itself \( y = x^3 \) to the line \( y = 8 \).
In summary, the region of integration helps us determine the limits for our double integral, defining where the integration starts and stops for both variables.

Order of Integration

The order of integration refers to the sequence of integrating variables in a double integral. It affects how the problem is approached.
In this problem, we decided to integrate \(y\) first followed by \(x\). Therefore, the integral is expressed as:

• First, integrate with respect to \( y \) from \( x^3 \) to \( 8 \).
• Then, integrate with respect to \( x \) from \( 0 \) to \( 2 \).

The choice often depends on the ease of evaluating integrals. Changing the order might simplify computations or make them feasible, especially if one variable has easier limits.

Evaluating Integrals

Evaluating integrals involves performing the integration process within specified limits. Let's break down:The inner integral is solved first, with respect to \( y \):\[\int_{x^3}^{8} 2xy\, dy = 2x \left[ \frac{1}{2} y^2 \right]_{x^3}^{8} = 2x \left( 32 - \frac{x^6}{2} \right) = 64x - x^7\]Here, we evaluated \( y \) from \( x^3 \) to \( 8 \).
Next, apply it to the outer integral with respect to \( x \):\[\int_{0}^{2} (64x - x^7)\, dx = \left[ 32x^2 - \frac{1}{8}x^8 \right]_{0}^{2}\]Calculating gives:\( 32 \times 4 - \frac{1}{8} \times 256 = 128 - 32 = 96 \).Thus, executing each integral carefully ensures comprehensive and accurate solutions.

Calculus

Calculus is the mathematical study of continuous change, and double integrals are a significant part of this discipline. They allow us to calculate areas, volumes, and accumulations over two-dimensional regions.
In this context, the integration of \( 2xy \) over a specific region reveals the accumulated value of that function within \( R \). It showcases the power of calculus in handling real-world phenomena by providing both theoretical and practical insights.
Double integrals serve as a foundation for many advanced topics in calculus, such as surface integrals and multivariable functions. Understanding these concepts is essential for exploring more complex systems and mathematical techniques.