Problem 12 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(7-20\) $$ \int_{-1}^{1} \int_{0}^{1} \int_{0}^{2}(x+y+z) d y d x d z $$

Step-by-Step Solution

Answer

The integral evaluates to 6.

Step 1: Evaluate the Innermost Integral

First, consider the innermost integral with respect to \( y \): \[ \int_{0}^{1} (x + y + z) \, dy \]Since \( x \) and \( z \) are treated as constants with respect to \( y \), apply the integral of a sum rule: \[ = \int_{0}^{1} x \, dy + \int_{0}^{1} y \, dy + \int_{0}^{1} z \, dy \] Evaluating these integrals gives: \[ = x[y]_{0}^{1} + \frac{1}{2}y^2\Bigg|_{0}^{1} + z[y]_{0}^{1} \] \[ = x(1 - 0) + \frac{1}{2}(1^2 - 0) + z(1 - 0) \]\[ = x + \frac{1}{2} + z \]

Step 2: Evaluate the Second Integral

Next, integrate the result from Step 1 with respect to \( x \):\[ \int_{-1}^{1} (x + \frac{1}{2} + z) \, dx \]Apply the sum rule:\[ = \int_{-1}^{1} x \, dx + \int_{-1}^{1} \frac{1}{2} \, dx + \int_{-1}^{1} z \, dx \]Evaluate these integrals: \[ = \frac{x^2}{2} \Bigg|_{-1}^{1} + \frac{1}{2}x \Bigg|_{-1}^{1} + z[x]_{-1}^{1} \]\[ = \left( \frac{1}{2}(1^2) - \frac{1}{2}((-1)^2) \right) + \frac{1}{2}(1 - (-1)) + z(1 - (-1)) \]\[ = (\frac{1}{2} - \frac{1}{2}) + 1 + 2z \]\[ = 1 + 2z \]

Step 3: Evaluate the Outermost Integral

Finally, integrate the result from Step 2 with respect to \( z \):\[ \int_{0}^{2} (1 + 2z) \, dz \]Apply the sum rule:\[ = \int_{0}^{2} 1 \, dz + \int_{0}^{2} 2z \, dz \] Evaluate these integrals: \[ = z \Bigg|_{0}^{2} + z^2 \Bigg|_{0}^{2} \] \[ = (2 - 0) + \left(2^2 \right) - \left(0^2 \right) \] \[ = 2 + 4 \] \[ = 6 \]

Step 4: Conclusion: Integral Result

The evaluated integral over the specified regions results in a value of \( 6 \). This completes the computation of the triple integral.

Key Concepts

Multiple Integration

Multiple integration is an extension of integration to functions of several variables. While single integration deals with the area under a curve for functions of one variable, multiple integration helps to find volumes and other multi-dimensional measures for functions involving two or more variables. In the given exercise, we're working with a triple integral, which is a type of multiple integration applied to three variables: \( x \), \( y \), and \( z \).A triple integral is typically expressed in nested form: one integral inside another. This allows us to compute the integral over a three-dimensional space by integrating in each variable's direction one at a time. This is often visually represented by considering layers of a volume.Here are some key points related to multiple integration:

• The order of integration can sometimes be rearranged for convenience, but it depends on how each integral's limits depend on the other variables.
• Multiple integration is crucial for calculating things like the mass of an object with variable density, or the charge in a non-uniform electric field.
• Triple integrals can be thought of as computing the accumulated "stuff" over a three-dimensional region, where "stuff" could be a scalar quantity like mass or charge density.

Iterated Integrals

Iterated integrals are a method for evaluating multiple integrals by breaking them down into a sequence of single integrals. In the context of our exercise, an iterated integral means evaluating the integral one variable at a time, starting from the innermost integral and moving outward.For the integral \( \int_{-1}^{1} \int_{0}^{1} \int_{0}^{2}(x+y+z) dy \, dx \, dz \):

• The first step is to integrate with respect to \( y \), treating \( x \) and \( z \) as constants. This simplifies the work as it reduces the complexity of handling all variables at once.
• The result from the innermost integral is then used as the function for the next integral with respect to \( x \).
• Finally, the outermost integral with respect to \( z \) gives the complete evaluation of the original iterated integral.

Iterated integrals take advantage of the fact that continuous integrals over separate variables can often be calculated independently, under certain conditions defined by Fubini's Theorem. This makes them a powerful tool in calculus, allowing for the simplification of complex multi-variable functions.

Calculus

Calculus is a branch of mathematics that studies continuous change. In the realm of multi-variable functions, calculus enables us to examine how these functions change in multi-dimensional space. Our discussion of triple integrals is deeply rooted in the principles of calculus, which provide the necessary tools to tackle such problems. Some basic but crucial concepts in calculus include:

• Differentiation: The process of finding the rate at which a function is changing at any given point. Its counterpart in multi-variable calculus is finding partial derivatives.
• Integration: The reverse process of differentiation, used to calculate things like area, volume, and total accumulation. It can be applied in more complex ways to functions with several variables, as seen with multiple integrals.
• The Fundamental Theorem of Calculus: This theorem provides a connection between differentiation and integration, playing a crucial role in calculating antiderivatives.

Understanding these principles allows us to utilize calculus effectively to model and solve real-world problems involving changes and areas in a wide variety of scientific fields, from physics to engineering.