Problem 10 - University Calculus: Early Transcendentals

Question

Evaluate the integrals in Exercises \(7-20\). $$\int_{0}^{1} \int_{0}^{3-3 x} \int_{0}^{3-3 x-y} d z d y d x$$

Step-by-Step Solution

Answer

The value of the integral is \(\frac{3}{2}\).

Step 1: Understand the Order of Integration

The given integral \( \int_{0}^{1} \int_{0}^{3-3x} \int_{0}^{3-3x-y} dz \ dy \ dx \) is a triple integral. This means we must integrate with respect to \(z\), then \(y\), and finally \(x\), in that order. It represents a volume under a function defined in a specific region of three-dimensional space.

Step 2: First Integration Over z

Integrate the innermost part of the triple integral with respect to \(z\):\[ \int_{0}^{3-3x-y} dz = [z]_{0}^{3-3x-y} = (3-3x-y) - 0 = 3-3x-y. \]This simplifies the triple integral to:\[ \int_{0}^{1} \int_{0}^{3-3x} (3-3x-y) \, dy \, dx. \]

Step 3: Second Integration Over y

Integrate the resulting expression with respect to \(y\):\[ \int_{0}^{3-3x} (3-3x-y) \, dy = \left[ (3-3x)y - \frac{y^2}{2} \right]_{0}^{3-3x}. \]Compute the limits of integration:\[(3-3x)(3-3x) - \frac{(3-3x)^2}{2} - 0 = (3-3x)^2 - \frac{(3-3x)^2}{2}. \]Simplify further:\[(3-3x)^2 \left(1 - \frac{1}{2}\right) = \frac{(3-3x)^2}{2}. \]This results in:\[ \int_{0}^{1} \frac{(3-3x)^2}{2} \, dx. \]

Step 4: Third Integration Over x

Now, integrate the expression with respect to \(x\):\[ \int_{0}^{1} \frac{(3-3x)^2}{2} \, dx \]First, expand \((3-3x)^2\) as:\[ (3-3x)^2 = (9 - 18x + 9x^2). \]Integrate term by term:\[ \frac{1}{2} \left( \int_{0}^{1} 9 \, dx - \int_{0}^{1} 18x \, dx + \int_{0}^{1} 9x^2 \, dx \right). \]Calculate each integral:- \( \int_{0}^{1} 9 \, dx = [9x]_{0}^{1} = 9. \)- \( \int_{0}^{1} 18x \, dx = [9x^2]_{0}^{1} = 9. \)- \( \int_{0}^{1} 9x^2 \, dx = [3x^3]_{0}^{1} = 3. \)Combine results:\[ \frac{1}{2} (9 - 9 + 3) = \frac{1}{2} \times 3 = \frac{3}{2}. \]

Step 5: Solution Conclusion

The value of the triple integral is \(\frac{3}{2}\). This represents the accumulated "volume" under the conditions specified by the integral limits.

Key Concepts

Integration by Parts

When working with integrals, a common technique used to simplify complex expressions is **Integration by Parts**. Although it is mainly applied to single-variable calculus, its principles underpin methods used in multivariable problems as well. To understand this concept, think of the product rule in reverse:\[ \int u \, dv = uv - \int v \, du \] where:

• \(u\) and \(dv\) are functions or expressions within the integral.
• \(v\) and \(du\) are their respective integrals and derivatives.

While the given exercise does not explicitly use integration by parts, understanding it helps in multivariable calculus where expressions need breaking down for simpler evaluations. Here, the importance lies in recognizing how chunks of expressions are integrated step by step. In single-variable terms, replacing complex products with simpler components aligns well with the step-by-step approach of solving the triple integral.

Volume Calculation

Finding volumes using triple integrals is a powerful method in calculus. The heart of this method is dividing a complex three-dimensional object into infinitesimally small parts and summing the contributions of these tiny volumes. Think of covering a complex shape with layers—from the bottom to the top. Here's a step-by-step breakdown of the process:- **Identify the Bounds**: Begin by identifying the limits of integration for each variable. These bounds form a box or region in space. In our exercise, the bounds are determined by the conditions: \(x\) from 0 to 1, \(y\) from 0 to \(3-3x\), and \(z\) from 0 to \(3-3x-y\).
- **Inner to Outer Integration**: Start with the innermost integral and proceed outwards. Each integration corresponds to summing over one dimension, ultimately concluding with the entire three-dimensional region.
- **Accumulating Volume**: Each level of integration adds depth to the calculation, effectively building the volume layer by layer from the innermost integral outward. In this specific problem, integrating \(z\), then \(y\), and finally \(x\) results in combining areas to achieve the total volume.
This calculated volume, represented by the final result \(\frac{3}{2}\), shows the methodical capturing of space within the specified region.

Multivariable Calculus

Multivariable calculus extends traditional calculus into higher dimensions, focusing on functions involving more than one variable. It offers tools to analyze the relationships, changes, and accumulations occurring in spaces that are beyond the usual two-dimensional scope.
In this particular problem, each step in solving the triple integral plays into the concepts of multivariable calculus:- **Understanding Multi-Dimensional Spaces**: The problem involves calculating volume within a 3D space defined by constraints and boundaries for \(x\), \(y\), and \(z\). Coordinate planes guide the integration path.
- **Iterated Integrals for Complexity Handling**: By slicing the problem into smaller parts with iterated integrals, each level breaks down the complex region into simpler, manageable steps. Multivariable calculus employs iterated integration frequently to handle problems involving shapes and volumes in spaces affected by multiple variables.
- **Real-World Applications**: These foundational concepts extend into physics, engineering, and other sciences where systems rely on multi-dimensional analysis for solving real-world challenges, like computing volumes, areas, or optimizing functions across various constraints.
The successful integration here demonstrates how multivariable calculus efficiently captures and calculates volumes, empowering problem-solving across diverse fields.