Problem 3 - Advanced Engineering Mathematics

Question

Evaluate the given iterated integral. $$ \int_{0}^{6} \int_{0}^{6-x} \int_{0}^{6-x-z} d y d z d x $$

Step-by-Step Solution

Answer

The value of the iterated integral is 36.

Step 1: Understanding the Problem

The integrals given are nested, or iterated, with respect to three variables: \(y\), \(z\), and \(x\). The limits of integration for \(y\) depend on \(z\) and \(x\), the limits for \(z\) depend on \(x\), and \(x\) goes from 0 to 6. Our task is to evaluate this integral step-by-step.

Step 2: Evaluate the Integral with Respect to y

First, evaluate the innermost integral, which is \(\int_{0}^{6-x-z} dy\). This integral is straightforward since \(dy\) has no variable dependencies, yielding the integral of 1 over the interval of integration. Therefore, \(\int_{0}^{6-x-z} dy = (6-x-z)\).

Step 3: Substitute the Result into the Middle Integral

Substitute the result of the first integral into the next integral: \(\int_{0}^{6-x}(6-x-z)dz\). Now we have to integrate \(6-x-z\) with respect to \(z\) from 0 to \(6-x\).

Step 4: Evaluate the Integral with Respect to z

Perform the integration: \(\int_{0}^{6-x}(6-x-z)dz = \left[(6-x)z - \frac{z^2}{2}\right]_{0}^{6-x}\). Substituting the limits gives \((6-x)(6-x) - \frac{(6-x)^2}{2} = (6-x)^2 - \frac{(6-x)^2}{2}\), which simplifies to \(\frac{(6-x)^2}{2}\).

Step 5: Substitute and Simplify into the Outer Integral

Substitute the result into the outermost integral: \(\int_{0}^{6} \frac{(6-x)^2}{2} dx\). Simplify this as \(\frac{1}{2} \int_{0}^{6} (6-x)^2 dx\).

Step 6: Evaluate the Final Integral with Respect to x

Perform the integration of \((6-x)^2\) with respect to \(x\):Let \(u = 6-x\) so that \(du = -dx\). Changing the limits accordingly: when \(x = 0\), \(u = 6\); and when \(x = 6\), \(u = 0\).Now substitute: \(\frac{1}{2} \int_{6}^{0} u^2 (-du) = \frac{1}{2} \int_{0}^{6} u^2 du\). This gives:\(\frac{1}{2} \left[ \frac{u^3}{3} \right]_{0}^{6} = \frac{1}{2} \left[ \frac{6^3}{3} - 0 \right] = \frac{1}{2} \times \frac{216}{3} = \frac{1}{2} \times 72 = 36\).

Key Concepts

Multivariable Calculus

In multivariable calculus, we deal with functions of several variables and their integrals. This type of calculus is particularly useful when we want to calculate areas, volumes, and more complex shapes beyond simple lines.
One of the primary tools of multivariable calculus is the iterated integral. This is an extension of single-variable calculus where we use multiple integrations to work across different dimensions.
This allows us to solve problems involving regions or shapes where the bounds are not simply constant numbers. Instead, they depend on other variables, requiring a careful setup and evaluation. In our example, the integral involves variables \(y\), \(z\), and \(x\), each requiring nested evaluation from the innermost to the outermost integral, reflecting a layered approach that is a hallmark of multivariable calculus.

Volume Calculation

Volume calculation using iterated integrals is a powerful method to find the volume of irregularly shaped regions, particularly in three dimensions. This is crucial when the region or object is complex and cannot be easily measured using simple geometric formulae.

• The given iterated integral \(\int_{0}^{6} \int_{0}^{6-x} \int_{0}^{6-x-z} d y d z d x\) represents a specific volume under certain constraints or bounds in three-dimensional space.
• The idea is to sum up all the tiny "slices" along each variable direction—essentially calculating the combined volume of these slices.
• Each integration step "accumulates" the volume under specific bounds set by other variables, akin to peeling layers from the inside out.

For students, understanding this approach is essential for extending their calculus toolkit from two dimensions to three or more.

Integration Techniques

In tackling iterated integrals, several integration techniques are applied to simplify and solve the problem.

• First, evaluate the integral from the innermost to the outermost. For instance, initially integrating with respect to \(y\), then \(z\), and finally \(x\), as in our exercise.
• A common technique used here is substitution, such as letting \(u = 6-x\) to transform the integral in a more convenient form to solve.
• Another important aspect is understanding how bounds change during integration, particularly how they depend on other variables in the problem, requiring careful manipulation to keep track of variable limits accurately.

By mastering these techniques, students can tackle a wide range of complex integrals in multivariable calculus, laying the groundwork for advanced applications in physics, engineering, and beyond.