Problem 23 - Thomas Calculus

Question

Find the volumes of the regions in Exercises \(23-36\) $$ \begin{array}{l}{\text { The region between the cylinder } z=y^{2} \text { and the } x y \text { -plane that is }} \\ {\text { bounded by the planes } x=0, x=1, y=-1, y=1}\end{array} $$

Step-by-Step Solution

Answer

The volume of the region is \(\frac{2}{3}\) cubic units.

Step 1: Understand the Problem

We need to find the volume of the region bounded by the cylinder \(z = y^2\) and the \(xy\)-plane, along with planes \(x = 0\), \(x = 1\), \(y = -1\), and \(y = 1\). This is a three-dimensional region in space.

Step 2: Set Up the Integral

Since the region is defined by \(x\), \(y\), and \(z\), we will use triple integration. Our limits of integration for \(x\) are from 0 to 1, for \(y\) from -1 to 1, and for \(z\) from 0 to \(y^2\). The volume \(V\) is given by:\[V = \int_{x=0}^{1} \int_{y=-1}^{1} \int_{z=0}^{y^2} dz \; dy \; dx\]

Step 3: Integrate with respect to \(z\)

First, integrate with respect to \(z\). Since the integral of 1 with respect to \(z\) is \(z\), the integration gives:\[\int_{z=0}^{y^2} dz = z \bigg|_{0}^{y^2} = y^2\]

Step 4: Integrate with respect to \(y\)

Now, substitute \(z = y^2\) into the integral and integrate with respect to \(y\):\[\int_{y=-1}^{1} y^2 \, dy\]Calculate this integral:\[\int y^2 \, dy = \frac{y^3}{3} \bigg|_{-1}^{1} = \left(\frac{1^3}{3} - \frac{(-1)^3}{3}\right) = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}\]

Step 5: Integrate with respect to \(x\)

Finally, integrate the result with respect to \(x\) from 0 to 1:\[\int_{0}^{1} \frac{2}{3} \; dx\]Since \(\frac{2}{3}\) is a constant, the integral becomes:\[\frac{2}{3}(x) \bigg|_{0}^{1} = \frac{2}{3} \times (1 - 0) = \frac{2}{3}\]

Step 6: Conclusion: Evaluate the Integral

The evaluated integral represents the volume of the region, which is \(\frac{2}{3}\) cubic units.

Key Concepts

Triple Integration

Triple integration is an extension of double integration into three dimensions. This mathematical tool is particularly useful for calculating volumes in three-dimensional space. In general terms, a triple integral is an expression:\[\int \int \int_{R} f(x, y, z) \; dx \; dy \; dz\]where \( R \) represents the region of integration in 3D space. In the context of this problem, we are exploring a volume confined by specific geometric boundaries. The triple integral essentially sums up small volumes (infinitesimal boxes) over the bounded region.

• Step 1: Break down the space into smaller elements, "cubes" or "boxes" based on chosen coordinate limits.
• Step 2: Integrate each portion with respect to one variable while keeping others fixed, moving consecutively from one variable to another.
• Step 3: Combine these calculations to determine the total volume or other desired property over the bounded region.

By understanding each integration's meaning, you can visualize the building up of the entire three-dimensional volume through continuous slices and layers.

Cylindrical Region

A cylindrical region in 3D space refers to areas that have a "cylinder-like" shape. In this exercise, the cylinder is described by the equation \(z = y^2\). This equation suggests that for every fixed \( y \), the \( z \)-value forms a parabola - resembling a cylinder vertically extending in the \( z \) direction.

• The cylinder has a square cross-section along the \(xy\)-plane, bounded by \(-1 \leq y \leq 1\) and \(x = 0\) to \(x = 1\).
• The height of the cylinder at any point is determined by \(z = y^2\), where \(z\) reaches zero when \(y = 0\).

This bounded region helps define spatial limits necessary for performing triple integration. Visualizing cylindrical regions makes it easier to comprehend the volume under examination, which in this problem, bears the shape of a parabolic cylinder.

Boundaries in 3D Space

Boundaries in 3D space define the limits for our integration. Without clear boundaries, it would be impossible to correctly calculate volumes or areas. For the region in this exercise, boundaries are explicitly set and must be closely adhered to when setting up integrals.

• \(x = 0\) to \(x = 1\) defines the span along the x-axis, acting like the length of the cylinder's base along the x-dimension.
• \(y = -1\) to \(y = 1\) gives the width of the base in the direction of the y-axis, forming another required boundary.
• The cylinder is capped by \(z = y^2\) and the \(xy\)-plane (\(z = 0\)), denoting the height limits of the cylindrical region.

Each boundary restricts the region in space where the volume is calculated, providing a clearer picture of the geometric area to be evaluated.

Integral Evaluation

When evaluating integrals, especially in triple integration, performing the calculations step-by-step aids in simplifying the mathematical process. Each step of integration isolates one variable, reducing complexity.

• Integrating with respect to \(z\) from 0 to \(y^2\), recognizing that the integral resolves to the bounds of \(z\) provides: \(z = y^2\).
• The next integration with respect to \(y\), \(\int_{-1}^{1} y^2 \, dy\), simplifies to \(\frac{y^3}{3}\) and evaluates to \(\frac{2}{3}\).
• Finally, integrating the constant \(\frac{2}{3}\) with respect to \(x\) from 0 to 1 is straightforward resulting in the product, \(\frac{2}{3}\).

Following these systematic steps ensures clear comprehension of how individual variable integrations contribute towards calculating the aggregate volume of the cylinder. The solution concludes at the volume, based on the given geometric constraints, offering precise understanding and application of integration.