Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the iterated integrals. \(\int_{-2}^{4} \int_{x-1}^{x+1} \int_{0}^{\sqrt{2 y / x}} 3 x y z d z d y d x\)

Step-by-Step Solution

Answer

The value of the iterated integral is 156.

Step 1: Evaluate the Innermost Integral

First, integrate with respect to \(z\). The innermost integral is \(\int_{0}^{\sqrt{2y/x}} 3xyz\, dz\).\ Since 3xy is a constant with respect to \(z\), we integrate \(z\) to get \(\frac{3xyz^2}{2}\) evaluated from \(0\) to \(\sqrt{2y/x}\).\[ \int_{0}^{\sqrt{2y/x}} 3xyz\, dz = \left[ \frac{3xyz^2}{2} \right]_{0}^{\sqrt{2y/x}} = \frac{3xy(2y/x)}{2} = 3y^2. \]

Step 2: Evaluate the Middle Integral

Next, integrate the result from Step 1 with respect to \(y\).\[ \int_{x-1}^{x+1} 3y^2\, dy. \] The antiderivative of \(3y^2\) is \(y^3\), so we evaluate it between \(x-1\) and \(x+1\):\[ \left[ y^3 \right]_{x-1}^{x+1} = (x+1)^3 - (x-1)^3. \]

Step 3: Simplify the Result of the Middle Integral

Simplify the expression \((x+1)^3 - (x-1)^3\): \[ (x+1)^3 = x^3 + 3x^2 + 3x + 1,\] \[ (x-1)^3 = x^3 - 3x^2 + 3x - 1. \] Thus, \( (x+1)^3 - (x-1)^3 = (x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1) = 6x^2 + 2.\)

Step 4: Evaluate the Outermost Integral

Finally, integrate the result \(6x^2 + 2\) with respect to \(x\) from \(-2\) to \(4\):\[ \int_{-2}^{4} (6x^2 + 2)\, dx. \] Compute the antiderivative: \(\int 6x^2\, dx = 2x^3\) and \( \int 2\, dx = 2x\). Evaluate the antiderivative from \(-2\) to \(4\):\[ \left[ 2x^3 + 2x \right]_{-2}^{4} = \left(2(4)^3 + 2(4)\right) - \left(2(-2)^3 + 2(-2)\right).\]

Step 5: Calculate the Final Result

Compute the above expression:\[ \left(2(4)^3 + 2(4)\right) = 2 \times 64 + 8 = 136. \] \[ \left(2(-2)^3 + 2(-2)\right) = 2 \times (-8) - 4 = -20. \] Thus, the final result is \(136 - (-20) = 136 + 20 = 156.\)

Key Concepts

Multiple Integration

Multiple integration extends the concept of single-variable integration to functions of multiple variables. When dealing with real-world problems, these functions usually depend on two or more variables.
For example, surfaces in a 3D space. Multiple integration helps compute volumes, areas, and other properties that involve multiple dimensions.
Iterated integrals are evaluations of these multiple integrations, and they involve performing the integration process in a step-by-step manner.

• Typically, we first deal with the innermost integral, then progress outward.
• Each step requires integration of the function with respect to a different variable, while treating other variables as constants.

In the given problem, the triple integral involves integration with respect to three variables: \(z\), \(y\), and \(x\), in that order.

Calculus

Calculus is a fundamental branch of mathematics that is used to understand changes between values related through functions. In the context of our problem, it provides the rules and notations that let us perform multiple integrals.
Calculus consists of two main branches:

• Differential calculus, which deals with the rate of change (derivatives).
• Integral calculus, which involves finding the total size or value, such as areas under curves (integrals).

In our problem, integral calculus is more relevant, as it allows us to find the volume under a surface described by the function \(3xyz\).
The iterated integral lets us break down this problem into sequential steps, each focusing on one variable at a time.

Definite Integral

A definite integral gives the accumulated sum of areas under or above a curve, bound by a specific range.
It gives a numerical value instead of a function, which makes it distinct from the indefinite integral, which describes a family of functions.
In our exercise, each step of integration with respect to \(z\), \(y\), and \(x\) is a definite integral with specific limits.

• The innermost integral limits for \(z\) are from \(0\) to \(\sqrt{2y/x}\).
• The middle integral limits for \(y\) range from \(x-1\) to \(x+1\).
• The outermost integral limits for \(x\) are from \(-2\) to \(4\).

Definite integrals provide the final value of 156 when we combine all steps.

Integration Techniques

Integration can sometimes be challenging, so multiple techniques have been developed to help in solving integrals.
These techniques include but are not limited to substitution, integration by parts, and partial fractions.
In our problem, the integration technique involved the sequential approach of evaluating iterated integrals.

• Each step simplifies to a basic integral, often easy to solve using well-known formulas.
• The calculation involves consistently plugging in the limits to evaluate the integral at each step.

Simplification skills are vital as seen in the steps where we simplified polynomial expressions.
This approach effectively breaks down complex integral problems into manageable parts.