Question

Evaluate the iterated integrals. \(\int_{0}^{5} \int_{-2}^{4} \int_{1}^{2} 6 x y^{2} z^{3} d x d y d z\)

Short answer

The value of the iterated integral is 33750.

Step-by-step solution

Understanding the Problem

We need to evaluate the iterated integral \( \int_{0}^{5} \int_{-2}^{4} \int_{1}^{2} 6x y^2 z^3 \ dx \ dy \ dz \). This means we integrate the function \( 6x y^2 z^3 \) first with respect to \( x \), then \( y \), and finally \( z \).

Integrate with Respect to x

The innermost integral is with respect to \( x \) from \( 1 \) to \( 2 \). We first integrate \( 6x y^2 z^3 \) with respect to \( x \): \[ \int_{1}^{2} 6x y^2 z^3 \ dx = \left[ 3x^2 y^2 z^3 \right]_{1}^{2} \] Substituting the limits of integration: \[ = 3(2)^2 y^2 z^3 - 3(1)^2 y^2 z^3 = 12 y^2 z^3 - 3 y^2 z^3 = 9 y^2 z^3 \].

Integrate with Respect to y

Now, integrate the result \( 9 y^2 z^3 \) with respect to \( y \) from \(-2\) to \(4\): \[ \int_{-2}^{4} 9y^2 z^3 \ dy = 9z^3 \left[ \frac{y^3}{3} \right]_{-2}^{4} \] Simplifying, we get: \[ 9z^3 \left( \frac{4^3}{3} - \frac{(-2)^3}{3} \right) = 9z^3 \left( \frac{64 + 8}{3} \right) = 9z^3 \cdot \frac{72}{3} = 216z^3 \].

Integrate with Respect to z

Finally, integrate \( 216z^3 \) with respect to \( z \) from \( 0 \) to \( 5 \): \[ \int_{0}^{5} 216z^3 \ dz = 216 \left[ \frac{z^4}{4} \right]_{0}^{5} \] Calculating this gives: \[ = 216 \left( \frac{5^4}{4} - \frac{0^4}{4} \right) = 216 \cdot \frac{625}{4} \] \[ = 216 \cdot 156.25 = 33750 \].

Final Answer

After evaluating the integrals in each respective order, we found the result of the entire iterated integral. Therefore, the value of \( \int_{0}^{5} \int_{-2}^{4} \int_{1}^{2} 6xy^2z^3 \ dx \ dy \ dz \) is \( 33750 \).

Key concepts

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions with more than one variable. It forms the foundational basis for exploring how functions behave in higher dimensions, such as in physics and engineering. In multivariable calculus, the idea of taking derivatives and integrals is extended to functions of two, three, or even more variables, which is essential for understanding complex systems.

Key concepts in multivariable calculus include:

• Partial Derivatives: These are derivatives of multivariable functions when all but one variable is kept constant.
• Multiple Integrals: These integrals involve more than one integration process and are frequently used in calculating volumes and other physical properties.
• Gradients and Vector Fields: These concepts help in determining the direction of maximum increase of a function and are used in optimization problems.

Understanding multivariable calculus is essential for evaluating problems in fields like fluid dynamics, electromagnetism, and any other domain involving multi-dimensional systems.

Integration

Integration is a fundamental operation in calculus that allows us to determine the accumulation of quantities, which could be areas under curves, volumes of solids, or other physical attributes. In the context of multivariable calculus, integration extends to functions of several variables, which involves integrating over regions in higher-dimensional spaces.

Key points to remember about integration:

• Indefinite vs Definite Integration: Indefinite integration results in a family of functions with an arbitrary constant, while definite integration evaluates the exact accumulation between two points.
• Techniques of Integration: Know various methods such as substitution, integration by parts, and partial fractions to tackle integrals of complex functions.
• Multiple Integration: Involving multiple iterations of integration, allowing you to address higher-dimensional problems.

Mastering integration techniques is crucial for solving advanced problems such as finding the center of mass, electrical charge distributions, and other applications requiring multi-factor calculations.

Triple Integrals

Triple integrals allow us to evaluate a function over a three-dimensional space, making them extremely useful for calculating volumes, mass distributions, and other properties in three dimensions. Unlike single and double integrals, triple integrals involve integrating across three different dimensions or layers, typically denoted by iteratively using three integral signs.

In the given problem, triple integrals are used to solve:

• Iterated Integration: Process of integrating the innermost function first, followed by the outer integrals with respect to subsequent variables.
• Order of Integration: This depends on the limits specified and the variable with respect to which you are integrating, emphasizing the need for understanding boundary considerations and variable relationships.
• Applications: Commonly used in geometry to determine the volume of an object and in physics to evaluate dynamics over a region.

Learning how to set up and evaluate triple integrals provides a significant skill in handling real-world applications involving three-dimensional spaces, such as computing the moment of inertia or analyzing fluid flow through a volume.