Problem 20 - Advanced Engineering Mathematics

Question

Sketch the region \(D\) whose volume \(V\) is given by the iterated integral. $$ \int_{1}^{3} \int_{0}^{1 / x} \int_{0}^{3} d y d z d x $$

Step-by-Step Solution

Answer

Region \(D\) is a 3D volume bounded by these conditions: \(1 \leq x \leq 3\), \(0 \leq y \leq \frac{1}{x}\), and \(0 \leq z \leq 3\), forming a shrinking sheet along \(x\).

Step 1: Understand the Boundaries for each Variable

The given integral is \(\int_{1}^{3} \int_{0}^{\frac{1}{x}} \int_{0}^{3} dy\, dz\, dx\). For the variable \(x\), the bounds are from 1 to 3. For the variable \(z\), it ranges from 0 to 3. Finally, for the variable \(y\), it ranges from 0 to \(\frac{1}{x}\). These bounds define the region \(D\) in the \(xyz\)-space.

Step 2: Visualize the Boundaries

For \(x\) ranging from 1 to 3, the region is considered along the \(x\)-axis. By graphing \(y = \frac{1}{x}\) against \(x\), which decreases from 1 when \(x=1\) to approximately \(0.33\) when \(x=3\), we find the upper boundary for \(y\). For each slice along \(x\), \(z\) can take any value from 0 to 3 independent of \(x\) and \(y\). This essentially forms a sheet for each particular \(x\).

Step 3: Sketch the Region in 3D

To sketch region \(D\), consider drawing a 3D box that starts from \(x=1\) to \(x=3\), \(z=0\) to \(z=3\), and is bounded above by the curve \(y = \frac{1}{x}\), which creates an upper boundary surface. For each \(x\), the region in the \(y\)-axis is a line from 0 up to \(\frac{1}{x}\), and \(z\) extends the surface vertically.

Step 4: Combine All Elements

Combine all the information: the \(y\)-axis shaping according to \(\frac{1}{x}\), the full length of \(z\) from 0 to 3, and the scope of \(x\) from 1 to 3. Hence, the region \(D\) appears as a solid that starts at \(x=1\) with a larger triangular cross-section that shrinks until \(x=3\).

Key Concepts

Volume Calculation

Calculating the volume of a region using an iterated integral is all about understanding how the bounds for each variable define a 3D region. Let's explore this using the example exercise where we are given an iterated integral:\[ \int_{1}^{3} \int_{0}^{1 / x} \int_{0}^{3} dy \ dz \ dx \]To find the volume, you repeat integrations over the limits for each variable. For the innermost integral, we integrate with respect to \(y\) from 0 to \(\frac{1}{x}\), keeping \(z\) and \(x\) constant. The result then is integrated with respect to \(z\) from 0 to 3, maintaining \(x\) constant. Finally, the total result is integrated with respect to \(x\) from 1 to 3. Each integration gives us the cumulative contribution of that variable dimension to the overall volume. This intricate setup allows us to account for layers upon layers of infinitesimally thin slices to obtain a complete volume.

3D Region Sketching

Sketching the 3D region defined by an iterated integral involves visualizing boundaries in the 3D space based on the provided limits. In our example, the bounds of integration provide a helpful map:

• \(x\) ranges from 1 to 3
• \(z\) ranges from 0 to 3
• \(y\) ranges from 0 to \(\frac{1}{x}\)

To visualize the region \(D\), start with the \(xz\)-plane by considering a slice along the \(x\)-axis. For each slice of \(x\), draw lines where \(z\) runs top to bottom from 0 to 3. Then look at the \(x-y\)-plane, drawing a curve where \(y = \frac{1}{x}\). This curve moves from \(y=1\) at \(x=1\) to \(y\approx 0.33\) at \(x=3\). Next, imagine stretching this 2D shape vertically up to \(z=3\) along the \(z\)-axis. As a result, you'll have a 3D shape that looks somewhat like a triangular prism that narrows with increasing \(x\). This approach aids in forming a mental picture of the region's extents and dimensions.

Integral Bounds Visualization

Visualizing the bounds in an iterated integral is crucial for understanding the full picture of the 3D region. Let's dive into how each bound functions:- **Inner Bound for \(y\)**: This limit, from 0 to \(\frac{1}{x}\), dynamically changes and specifies that \(y\) gets smaller as \(x\) grows, forming an upper curve that's boundary-shaped.- **Middle Bound for \(z\)**: This limit remains static from 0 to 3, indicating a consistent height of the region without regard to the \(x\) and \(y\) dimensions. Imagine it as a series of vertical sheets.- **Outer Bound for \(x\)**: This limit, from 1 to 3, places outer constraints on the length of the region. The blocks defined by \(x\) stack side by side.By picturing these bounds step-by-step, you understand the exciting way each variable contributes to the region's overall shape and utilization in space. This skill becomes immensely useful in identifying the shape volume and making complex integrations manageable.