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 evaluated integral is \(\frac{3}{2}\).

Step 1: Analyze the Threefold Integral

We have a triple integral to evaluate: \[ \int_{0}^{1} \int_{0}^{3-3x} \int_{0}^{3-3x-y} dz \, dy \, dx \]This integral's order of integration is \(dz\), \(dy\), and \(dx\). The limits of integration in the reverse order define a region in space.

Step 2: Integrate with Respect to z

The inner integral is with respect to \(z\), which has constant limits from \(0\) to \(3-3x-y\).Perform the integration:\[ \int_{0}^{3-3x-y} dz = \left[ z \right]_{0}^{3-3x-y} = 3-3x-y.\]Substitute the result back into the integral for \(y\) and \(x\).

Step 3: Integrate with Respect to y

Now integrate the expression \(3-3x-y\) with respect to \(y\) from \(0\) to \(3-3x\):\[ \int_{0}^{3-3x} (3-3x-y) \, dy.\]Split and integrate each term individually:\[ \int_{0}^{3-3x} 3 \, dy = 3y \bigg|_{0}^{3-3x} = 9 - 9x, \]\[ \int_{0}^{3-3x} (-3x) \, dy = -3xy \bigg|_{0}^{3-3x} = -3x(3-3x), \]\[ \int_{0}^{3-3x} (-y) \, dy = -\frac{1}{2}y^2 \bigg|_{0}^{3-3x} = -\frac{1}{2}(3-3x)^2. \]

Step 4: Simplify y-integration Results and Finalize x-Integration

Combine the results from Step 3:\[ (9 - 9x) + [-3x(3 - 3x)] + \left[ -\frac{1}{2}(3 - 3x)^2 \right].\]Simplify the polynomial expressions and evaluate for \(x\):1. \[-3x(3-3x) = -9x + 9x^2\]2. \[-\frac{1}{2}(9 - 18x + 9x^2) = -\frac{9}{2} + 9x - \frac{9}{2}x^2\]Thus, the new integral becomes:\[ \int_{0}^{1} \left( 3 - 9x + 9x^2 \right) dx.\]

Step 5: Integrate with Respect to x

The final integration is straightforward:\[ \int_{0}^{1} (3-9x+9x^2) dx.\]Integrating term by term gives:\[ 3x - \frac{9}{2}x^2 + 3x^3 \bigg|_{0}^{1} = (3 - \frac{9}{2} + 3) - (0) = 6 - \frac{9}{2}.\]This simplifies to \[\frac{3}{2}.\]

Step 6: Verify and Conclude

Ensure calculations were correct by back-checking each integral's evaluations and simplification steps. Given the calculations, the volume under the described limits is correctly evaluated. Therefore, \[ \boxed{\frac{3}{2}} \] is the final answer.

Key Concepts

Iterated Integrals

Triple integrals are used to evaluate the volume under a surface within a three-dimensional region. In this specific problem, an ordered approach known as iterated integration is applied. We start by integrating with respect to one variable, then proceed with the other variables in sequence.

• Here, the integral's order is \(dz, dy, dx\).
• We solve step by step, starting with the innermost integral and moving outward.

This method simplifies complex calculations, as it breaks down the problem into more manageable parts. Performing the integration one layer at a time ensures that the various elements of the region are accurately computed.
Once the inner integral over \(z\) is done, its result is then used in the next integral over \(y\), followed by \(x\). Iterated integrals reveal the structure of the region being integrated over, allowing for systematic evaluation of multi-dimensional shapes.

Volume Calculation

When dealing with three-dimensional problems, such as this one, the objective is often to determine the volume bounded by certain surfaces and planes. The given triple integral represents the calculation of this enclosed space.

Breaking Down the Process:

• The integral's bounds \( (0 \leq z \leq 3-3x-y), (0 \leq y \leq 3-3x), (0 \leq x \leq 1) \) hint at a specific geometric shape.
• By integrating over \(z\), we calculate the height of the volume slice at each point \((x, y)\).
• The subsequent integration over \(y\) and \(x\) layers these slices into the total volume.

The ultimate result from such integrals gives the measure of space enveloped by the described region. In this example, through careful integration and simplification, we find the total volume to be \( \frac{3}{2} \), capturing how much space the region takes within its bounds.

Region of Integration

Understanding the region of integration is essential in solving triple integrals as they define the limits of the integration process. These limits determine the shape and dimensions of the area you're evaluating. Let's dissect this further:

Visualizing the Region:

• The order \( (0 \leq z \leq 3-3x-y), (0 \leq y \leq 3-3x), (0 \leq x \leq 1) \) constructs a three-dimensional right-angled region or a pyramid-like area in the xyz-space.
• The boundaries suggest a volume that's gradually tapering, emphasizing the dependence of each upper limit on prior variables.
• The integral stacks these limits from \(x\) through \(z\), giving a clear path through the volume to be calculated.

Deciphering the interdependent relationships between these limits is a crucial aspect of understanding the physical interpretation of the integral. Each boundary acts like a slice through the 3D space, pinpointing exactly where the measures of interest lie. This clarity helps make sense of the volume evaluation process, ensuring accurate outcomes.