Problem 10 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(7-20\) $$ \int_{0}^{1} \int_{0}^{3-3 x} \int_{0}^{3-3 x-y} d z d y d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the integral limits and order

The integral is presented as a triple integral: \[\int_{0}^{1} \int_{0}^{3-3x} \int_{0}^{3-3x-y} dz \, dy \, dx\]This means we first integrate with respect to \(z\), then \(y\), and finally \(x\). The integration limits are specified: for \(z\) it goes from 0 to \(3-3x-y\), for \(y\) from 0 to \(3-3x\), and for \(x\) from 0 to 1.

Step 2: Integrate with respect to z

Perform the integration of the innermost integral with respect to \(z\): \[\int_{0}^{3-3x-y} dz = z \bigg|_{0}^{3-3x-y} = (3-3x-y) - 0 = 3-3x-y\]Substitute the result in the remaining integral.

Step 3: Integrate with respect to y

Substitute \(3-3x-y\) into the integral with respect to \(y\):\[\int_{0}^{3-3x} (3-3x-y) \, dy\]To integrate, treat \(3-3x\) as a constant:\[\int (3-3x) \, dy - \int y \, dy\]Integrating gives:\[ (3-3x)y - \frac{y^2}{2} \bigg|_{0}^{3-3x}\]Plug in the limits:\[ (3-3x)(3-3x) - \frac{(3-3x)^2}{2}\]Simplifying:\[ 9 - 18x + 9x^2 - \frac{9 - 18x + 9x^2}{2} = \frac{9 - 18x + 9x^2}{2}\]

Step 4: Integrate with respect to x

Now integrate the expression with respect to \(x\):\[\int_{0}^{1} \frac{9 - 18x + 9x^2}{2} \, dx\]Distribute the integral:\[\frac{1}{2} \left[ 9x - 9x^2 + \frac{9x^3}{3} \right]_{0}^{1} \]Plugging in the limits gives:\[\frac{1}{2} \left[ 9(1) - 9(1)^2 + 3(1)^3 - (0 - 0 + 0) \right] = \frac{1}{2} [9 - 9 + 3] = \frac{3}{2}\]

Step 5: Final Answer

The entire triple integral has been evaluated, and the final result is:\[\frac{3}{2}\]

Key Concepts

Limits of Integration

In the context of a triple integral, the limits of integration play a crucial role. They define the region over which we are integrating the function. By specifying a range for each variable, they help us understand the boundaries of the three-dimensional region in space.

For the given integral:

• The inner limit is for the variable \(z\), ranging from \(0\) to \(3-3x-y\). This suggests that the upper boundary of \(z\) changes as \(x\) and \(y\) vary.
• The middle limit is for the variable \(y\), ranging from \(0\) to \(3-3x\). The upper boundary here is dependent on \(x\) but independent of \(z\).
• The outer limit is for the variable \(x\), running from \(0\) to \(1\), simple number limits that remain constant.

The integration proceeds from the innermost variable to the outermost, respecting these limits, allowing us to cover the defined volume inside the bounds.

Order of Integration

The order of integration dictates the sequence of variables to be integrated in a multiple integral. In a triple integral, there are six possible orders to follow, but the configurations of limits often hint at the simplest path.

For the exercise at hand, we are instructed to integrate with respect to \(z\) first, followed by \(y\), and finally \(x\). This order simplifies the computation as:

• Integrating \(z\) first allows us to deal with its direct limits \(0\) to \(3 - 3x - y\), reducing the function step-by-step.
• Once \(z\) is eliminated, integrating over \(y\) next uses its limits \(0\) to \(3-3x\), which are straightforward after the \(z\) integration.
• Finally, integrating over \(x\) ties it all together with its straightforward numeric limits.

Such a strategic ordering minimizes complexity and ensures that the function obtained after each step is manageable.

Multivariable Calculus

Multivariable Calculus expands the calculus of single variables to functions involving several variables. It allows us to explore scenarios where quantities depend on more than one factor. In this exercise, the triple integral signifies a fundamental concept of multivariable calculus. It generalizes the idea of integration from a line (one dimension) or a surface (two dimensions) to a volume (three dimensions).

With triple integrals, we can calculate volumes, mass, and other properties that arise when considering three-dimensional space. Each integration step effectively "slices" the volume down to simpler parts:

• First, we resolve the innermost integration by assuming the other variables are constants.
• Subsequently, simplifying and integrating out each variable allows us to tackle the entire three-dimensional area, layer by-layer.

This approach illustrates how multivariable calculus enables the analysis and understanding of more complex systems and spaces beyond single-variable calculus.