Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

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

Step 1: Integrate with respect to y

The given integral is \( \int_{0}^{2} \int_{1}^{z} \int_{0}^{\sqrt{x / z}} 2 x y z \, dy \, dx \, dz \). First, we will integrate \( 2 x y z \) with respect to \( y \) over the interval from \( 0 \) to \( \sqrt{x / z} \). The antiderivative of \( 2 x y z \) with respect to \( y \) is \( x y^2 z \). Evaluating at the bounds gives:\[ \left. x y^2 z \right|_{0}^{\sqrt{x/z}} = x (\sqrt{x/z})^2 z - x (0)^2 z = x \left(\frac{x}{z}\right) z = x^2. \]

Step 2: Integrate with respect to x

Next, integrate the result, \( x^2 \), with respect to \( x \) over the interval from \( 1 \) to \( z \):\[ \int_{1}^{z} x^2 \, dx. \]The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \). Evaluating at the bounds gives:\[ \left. \frac{x^3}{3} \right|_{1}^{z} = \frac{z^3}{3} - \frac{1^3}{3} = \frac{z^3 - 1}{3}. \]

Step 3: Integrate with respect to z

Finally, integrate the expression \( \frac{z^3 - 1}{3} \) with respect to \( z \) over the interval from \( 0 \) to \( 2 \):\[ \int_{0}^{2} \frac{z^3 - 1}{3} \, dz. \]We can split this into two separate integrals:1. \( \frac{1}{3} \int_{0}^{2} z^3 \, dz \)2. \( -\frac{1}{3} \int_{0}^{2} 1 \, dz \)The antiderivative of \( z^3 \) is \( \frac{z^4}{4} \), and of \( 1 \) is \( z \). Evaluating these gives:\( \frac{1}{3} \left[ \left. \frac{z^4}{4} \right|_0^2 \right] = \frac{1}{3} \left( \frac{16}{4} - 0 \right) = \frac{4}{3} \)\( -\frac{1}{3} \left[ \left. z \right|_0^2 \right] = -\frac{1}{3} \left( 2 - 0 \right) = -\frac{2}{3} \)Combining these results:\[ \frac{4}{3} - \frac{2}{3} = \frac{2}{3}. \]

Key Concepts

Multiple Integration

Multiple integration is a fundamental concept in calculus that extends the idea of a single integral to more than one dimension. While a single integral operates over one variable, multiple integrals allow us to compute volumes and accumulations over a region in two or more dimensions.

When you encounter iterated integrals, like in the original exercise, you're dealing with multiple layers of integration:

• The innermost integral evaluates first, focusing on one variable.
• Each successive integration considers the results over increasingly broad variables.

This method is often used to find volumes under surfaces or evaluate more complex regions. The process of approaching these problems involves breaking down the problem into simpler iterated integral steps. Each step handles one dimension of integration, just as shown in the given problem working through \(y\), \(x\), and \(z\) in succession.

Antiderivative

The term antiderivative is one you’ll frequently encounter in integration problems. It represents the inverse operation of taking a derivative. When we perform integration, we look for a function whose derivative yields the function we started with.

In the original exercise, the antiderivative of mixed variables was crucial at each step:

• For integrating \(2 x y z\) with respect to \(y\), the antiderivative is \(x y^2 z\).
• When tackling \(x^2\) with respect to \(x\), it becomes \(\frac{x^3}{3}\).
• Finally, the antiderivative of \(z^3 - 1\) with respect to \(z\) breaks down into simpler forms: \(-\frac{z^4}{4}\) for the \(z^3\) term and \(z\) for the constant.

Antiderivatives are a building block for evaluating definite integrals, as they allow us to use the Fundamental Theorem of Calculus, relating derivatives to integrals.

Definite Integral

A definite integral helps in finding the exact value of the accumulation of quantities, like area under a curve, over a specific interval. It's evaluated using limits as boundaries to provide a precise result.

The original problem demonstrated the use of definite integrals in three stages:

• First, for \(y\) from \(0\) to \(\sqrt{x/z}\), producing an expression solely in terms of \(x\) and \(z\).
• Second, integrating with respect to \(x\) between \(1\) and \(z\) further simplifies the problem.
• Lastly, the integration within \(z\) from \(0\) to \(2\) provided the final value of \(\frac{2}{3}\).

Definite integrals can also be broken down into multiple smaller integrals for more complex equations, as shown with the final separation between \(\frac{z^3}{4}\) and \(-1\). The solution involves evaluating these antiderivatives at the bounds of each variable's range.

Calculus

Calculus is a branch of mathematics that studies continuous change. It's essentially divided into two main areas: differential calculus, which deals with rates of change and slopes, and integral calculus, which focuses on accumulations, areas, and volumes.

In tackling iterated integrals, calculus connects these two concepts through the Fundamental Theorem of Calculus:

• Differentiation and integration are inverse operations.
• An integral can yield the net accumulation of quantities over a range.

The original exercise showcases the power of calculus in systematically solving complex mathematical problems involving multiple variables and dimensions. Understanding each part of calculus will solidify your problem-solving skills, opening up new ways to approach mathematical concepts both in academic and real-world applications.