Problem 2 - Advanced Engineering Mathematics

Question

Evaluate the given iterated integral. $$ \int_{1}^{3} \int_{1}^{x} \int_{2}^{x y} 24 x y d z d y d x $$

Step-by-Step Solution

Answer

The value of the iterated integral is approximately 374.67.

Step 1: Understand the Order of Integration

The integral is given as \( \int_{1}^{3} \int_{1}^{x} \int_{2}^{x y} 24 x y \, d z \, d y \, d x \). This is an iterated integral, which means we integrate from the innermost to the outermost. Here, we integrate over \(z\) first, then \(y\), and finally \(x\).

Step 2: Integrate with Respect to z

The innermost integral \( \int_{2}^{x y} 24 x y \, d z \) is with respect to \(z\). Since \(24 x y\) is independent of \(z\), it integrates to \(24 x y (z)\) evaluated from \(z = 2\) to \(z = xy\). This results in the expression: \(24 x y (x y - 2)\).

Step 3: Simplify After First Integration

After evaluating the innermost integral, we simplify the expression: \(24 x^2 y^2 - 48 x y\). Now, the iterated integral becomes \(\int_{1}^{3} \int_{1}^{x} (24 x^2 y^2 - 48 x y) \, d y \, d x\).

Step 4: Integrate with Respect to y

Now, integrate \(24 x^2 y^2 - 48 x y\) with respect to \(y\):\[\int_{1}^{x} (24 x^2 y^2 - 48 x y) \, d y = \int_{1}^{x} 24 x^2 y^2 \, d y - \int_{1}^{x} 48 x y \, d y\].The integrals are:1. \(\int_{1}^{x} 24 x^2 y^2 \, d y = 24 x^2 \left[ \frac{y^3}{3} \right]_{1}^{x} = 24 x^2 \left( \frac{x^3}{3} - \frac{1}{3} \right) = 8 x^5 - 8 x^2\).2. \(\int_{1}^{x} 48 x y \, d y = 48 x \left[ \frac{y^2}{2} \right]_{1}^{x} = 48 x \left( \frac{x^2}{2} - \frac{1}{2} \right) = 24 x^3 - 24 x\).

Step 5: Simplify After Second Integration

Combine the results of the two components from Step 4: \((8 x^5 - 8 x^2) - (24 x^3 - 24 x) = 8 x^5 - 24 x^3 - 8 x^2 + 24 x\). So, the integral becomes \(\int_{1}^{3} (8 x^5 - 24 x^3 - 8 x^2 + 24 x) \, d x\).

Step 6: Integrate with Respect to x

Integrate each term with respect to \(x\):1. \(\int 8 x^5 \, d x = \frac{8 x^6}{6} = \frac{4 x^6}{3}\).2. \(\int -24 x^3 \, d x = -\frac{24 x^4}{4} = -6 x^4\).3. \(\int -8 x^2 \, d x = -\frac{8 x^3}{3}\).4. \(\int 24 x \, d x = 12 x^2\).Combine these and evaluate from 1 to 3.

Step 7: Evaluate from 1 to 3

Substitute 3 and 1 into the resulting expression:\[\left[\frac{4x^6}{3} - 6x^4 - \frac{8x^3}{3} + 12x^2\right]_1^3\].Evaluating at \(x=3\):\(\frac{4(3)^6}{3} - 6(3)^4 - \frac{8(3)^3}{3} + 12(3)^2\) = \( 972 - 486 - 216 + 108 = 378 \).Evaluating at \(x=1\):\(\frac{4(1)^6}{3} - 6(1)^4 - \frac{8(1)^3}{3} + 12(1)^2\) = \( \frac{4}{3} - 6 - \frac{8}{3} + 12 = \frac{10}{3} \).Now subtract: \(378 - \frac{10}{3}\).

Step 8: Simplify the Result

The simplification provides \(378 - \frac{10}{3} = \frac{1134}{3} - \frac{10}{3} = \frac{1124}{3}\). Simplifying gives approximately \(374.67\).

Key Concepts

Multiple Integrals

Multiple integrals are an essential part of calculus that allow us to compute volumes, masses, and other quantities that accumulate over areas or volumes. They extend beyond single integrals, which only involve one variable, and instead, involve integrating functions of more than one variable over a specific region.
In the context of this exercise, we worked with a triple iterated integral, which means we computed the integration in three successive stages. Each stage corresponds to a variable: starting with the innermost integral and progressing outward.

• The innermost integral is \(\int_{2}^{xy} 24 xy \, d z\) and it handles the accumulation along the variable \(z\).
• Next, the intermediate integral \(\int_{1}^{x} \, d y\) looks into the changes along \(y\), based on the result found from the \(z\) integration.
• Finally, \(\int_{1}^{3} \, d x\) encompasses the outermost integration, summing everything up as it accounts for the variations in \(x\).

Multiple integrals are indispensable tools in fields ranging from physics to engineering, providing a mathematical lens to observe how quantities spread and evolve in multi-dimensional space.

Order of Integration

Grasping the order of integration is crucial when working with iterated integrals. It tells us the sequence in which we should carry out the integration with respect to each variable.
In our example, we have the order as \(d z \, d y \, d x\). Essentially, this means:

• Start by integrating with respect to \(z\). In our problem, \(z\) varies from 2 to \(xy\).
• Proceed by tackling \(y\), which varies from 1 to \(x\), as specified in the exercise.
• Finally, move onto \(x\), which spans from 1 to 3.

This clear order is essential, as changing the order of integration can significantly alter the steps needed and the resulting outcome.
When tackling such problems, carefully determine the limits of each integration level and adjust them according to the variable and its relative position in the sequence.

Techniques of Integration

The techniques of integration employed here mainly involve straightforward substitution and algebraic manipulation. Understanding these techniques can ease solving even the most complex integrals.
In our solved exercise:

• We used basic algebra to integrate with respect to \(z\), leveraging that 24xy is independent of \(z\), allowing simple evaluation over its limits.
• Next, integration with respect to \(y\) called for breaking the integral into manageable parts: \(\int 24 x^2 y^2 \, d y\) and \(\int 48 x y \, d y\). We used power rule for integration to progress through these parts smoothly.
• For integration over \(x\), again we utilized substitution followed by evaluation over the defined limits, integrating each polynomial expression representing different components.

Familiarity with these techniques will grant you the ability to handle complex integrals by reducing them to straightforward calculations. It highlights the importance of a strong foundation in calculus concepts to simplify and solve these multi-layered problems effectively.