Problem 12 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Understand the Problem

We are asked to evaluate the triple integral \( \int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} (x+y+z)\, dy\, dx\, dz \). This means we need to integrate the function \( x+y+z \) over a cube defined by \( -1 \leq x, y, z \leq 1 \).

Step 2: Evaluate the Inner Integral with Respect to y

We first integrate the function \( x+y+z \) with respect to \( y \) from \( -1 \) to \( 1 \): \[ \int_{-1}^{1} (x + y + z) \, dy = \left[ xy + \frac{y^2}{2} + yz \right]_{-1}^{1} = \left( x(1) + \frac{1^2}{2} + z(1) \right) - \left( x(-1) + \frac{(-1)^2}{2} + z(-1) \right) \]Simplify further: \[ = (x + \frac{1}{2} + z) - (-x + \frac{1}{2} - z) = 2x + 2z \]

Step 3: Evaluate the Middle Integral with Respect to x

Next, integrate \( 2x + 2z \) with respect to \( x \) from \( -1 \) to \( 1 \):\[ \int_{-1}^{1} (2x + 2z) \, dx = \left[ x^2 + 2zx \right]_{-1}^{1} = \left( 1^2 + 2z(1) \right) - \left( (-1)^2 + 2z(-1) \right) \]Simplify:\[ = (1 + 2z) - (1 - 2z) = 4z \]

Step 4: Evaluate the Outer Integral with Respect to z

Finally, integrate \( 4z \) with respect to \( z \) from \( -1 \) to \( 1 \):\[ \int_{-1}^{1} 4z \, dz = \left[ 2z^2 \right]_{-1}^{1} = 2(1^2) - 2((-1)^2) = 2 - 2 = 0 \]

Step 5: Conclusion

The value of the triple integral \( \int_{-1}^{1} \int_{-1}^{1} \int_{-1}^{1} (x+y+z) \, dy \, dx \, dz \) over the specified limits is 0.

Key Concepts

Iterated Integrals

Iterated integrals are a method used to evaluate multiple integrals by performing integration one variable at a time in a nested sequence. In the given problem, we have a triple integral which involves integrating with respect to three variables:\( x, y, \, \text{and} \, z \). This means we perform an integral with respect to one variable while treating the others as constants, and repeat the process for each variable.
Here's the order of integration:

• First, integrate with respect to \( y \)
• Second, integrate the resulting function with respect to \( x \)
• Finally, integrate the resulting expression with respect to \( z \)

This systematic approach allows us to simplify a complex integral by breaking it into manageable steps. Understanding the order in iterated integrals is essential to accurately solve problems involving more than one variable.

Integration Over a Cube

When integrating over a cube, it's important to consider the region of integration and the limits associated with each variable. The cube referred to in this exercise has boundaries defined by the limits \(-1 \leq x, y, z \leq 1\). This means our domain of integration is a perfect cube centered at the origin with side length of 2 units.
In practical terms, we are integrating the function \( x+y+z \) over all points within this cube. This provides a well-defined and bounded three-dimensional space, simplifying the evaluation process since all variables share the same symmetric limits.
Understanding the geometry and limits aids in effectively setting up and solving the triple integral.

Evaluation of Definite Integrals

Evaluating definite integrals involves calculating the "accumulated" value of a function over an interval, providing a clear result rather than an indefinite answer with constants.
For each step in our iterated integration:

• After integrating with respect to \( y \): The result is \( 2x + 2z \).
• We then integrate with respect to \( x \): This yields \( 4z \).
• Finally, integrating with respect to \( z \) results in the ultimate value of \( 0 \).

In each step, we substitute the upper and lower limits for the variable of integration, applying the Fundamental Theorem of Calculus. This method provides a definitive solution, allowing you to calculate the exact value to be evaluated across the specified range.

Symmetric Limits

Symmetric limits in integration refer to limits that are equidistant from a central value, often simplifying calculations due to symmetry. In this problem, each variable \( x, y, \text{and } z\) ranges from \(-1\) to \(1\).
Such symmetry offers benefits, including:

• Identifying even or odd functions. Here, \( x+y+z \) is odd over the symmetric limits. This symmetry can sometimes lead integrals, like in this case, to become zero.
• Reducing computational complexity by providing consistent bounds for each variable.
• Highlighting the central point (\(0\) in this case), allowing for simplifications in multivariable calculus when integrating functions evenly distributed about this point.

Recognizing and leveraging symmetric limits speeds up and simplifies the integration process.