Problem 23 - Thomas Calculus

Question

The region between the cylinder \(z=y^{2}\) and the \(x y\) -plane that is bounded by the planes \(x=0, x=1, y=-1, y=1\) (GRAPH NOT COPY)

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We are tasked with finding the volume of the region that is between the paraboloid given by the equation \( z = y^2 \) and the \( xy \)-plane. This region is bounded by the planes \( x=0 \), \( x=1 \), \( y=-1 \), and \( y=1 \). This means our region lies in the first cylinder with boundaries for \( x \) and \( y \).

Step 2: Set Up the Double Integral

The volume can be found by integrating the function \( z = y^2 \) over the rectangular region in the \( xy \)-plane defined by the boundaries. The double integral becomes:\[ V = \int_{x=0}^{1} \int_{y=-1}^{1} y^2 \, dy \, dx\]

Step 3: Integrate with Respect to y

First, integrate \( y^2 \) with respect to \( y \) over \( y = -1 \) to \( y = 1 \):\[\int_{-1}^{1} y^2 \, dy = \left[ \frac{y^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \left( \frac{(-1)^3}{3} \right) = \frac{1}{3} - \left( -\frac{1}{3} \right) = \frac{2}{3}\]

Step 4: Integrate with Respect to x

Now, integrate the result of the previous integral with respect to \( x \) over the interval \( x = 0 \) to \( x = 1 \):\[\int_{0}^{1} \frac{2}{3} \, dx = \frac{2}{3} \cdot (1-0) = \frac{2}{3}\]

Step 5: Conclusion

The volume of the region between the paraboloid \( z = y^2 \) and the \( xy \)-plane, bounded by the planes \( x = 0 \), \( x = 1 \), \( y = -1 \), and \( y = 1 \), is \( \frac{2}{3} \) cubic units.

Key Concepts

Volume Calculation Using Double Integrals

To find the volume of a region in three-dimensional space, one very effective technique is using double integrals. The double integral allows us to sum up infinitesimally small volumes in a given region, calculating a total volume. Imagine slicing the region into thin, flat layers parallel to the base area. We sum the volume of these slices to find the total volume.

The process starts by defining the boundaries of the region, which in most cases is a rectangular area in the base plane, known as the domain. This is done using the limits of integration for the double integral. The function that needs to be integrated, often referred to as the 'height' function, is placed over this domain. In the case of the exercise, it's the function representing the paraboloid, which is given by the equation: \[V = \int_{x_0}^{x_1} \int_{y_0}^{y_1} f(x, y) \, dy \, dx\]

• First, integrate 'f(x, y)' with respect to 'y', from 'y_0' to 'y_1'.
• Then, take the result of this integration and integrate with respect to 'x', from 'x_0' to 'x_1'.

The result of these steps gives the volume of the region beneath the surface and above the given domain in the base plane.

Understanding Bounded Regions in Double Integrals

Bounded regions in the plane provide the limits needed for double integrals. Visualizing these boundaries helps when setting up the integral since they determine the area over which you'll integrate the function.

In this exercise, the region is described by linear constraints:

• \(x = 0\) to \(x = 1\)
• \(y = -1\) to \(y = 1\)

This forms a rectangular area on the \(xy\)-plane. Within these bounds, the function \(z=y^2\) is integrated. This specific bounded region makes solving the double integral straightforward since the limits are constant. Such limits lead to simple rectangular domains.

Bounding regions are crucial as they confine the volume of interest. They simplify our computations by creating a fixed plane area where our 3D surface, represented by the function, interacts with the plane.

Exploring Paraboloids in Three-Dimensional Space

Paraboloids are three-dimensional surfaces that resemble an upward-opening or downward-opening dish. They are a secondary category of quadric surfaces. Every point on a paraboloid is equidistant from a fixed point (called the focus) and a fixed line (called the directrix).

The equation given in the exercise, \(z = y^2\), represents an elliptic paraboloid. This surface opens upwards along the \(z\)-axis. If we visualize it, it's as though the surface is curving upwards above the \(xy\)-plane.

• The "y" variable in the equation gives a parabolic curve when viewed in any plane parallel to the \(yz\)-plane.
• "z" acts as both a height and a function of "y", showing how paraboloids extend into three-dimensional space from flat 2D curves.

Paraboloids are significant in many areas of study, including physics and engineering, due to their natural occurrence in phenomena like gravity wells and reflecting surfaces. Understanding them helps in applying calculus concepts like integration to find volumes and areas involving these surfaces.