Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the given solid. First, sketch the solid; then estimate its volume; finally, determine its exact volume. Solid in the first octant enclosed by \(z=4-x^{2}\) and \(y=2\)

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{128}{3}\) cubic units.

Step 1: Sketch the Solid

The given solid lies in the first octant and is enclosed by the surface defined by the equation \(z=4-x^{2}\) and the plane \(y=2\). This surface represents a parabolic cylinder that opens downward with vertex at \((0,0,4)\). The restriction \(y=2\) means that the solid is bounded between \(y=0\) and \(y=2\). In the \(xy\)-plane, the parabola \(z=4-x^{2}\) traces out a curve from \(x=-2\) to \(x=2\). Hence, the region of integration in the xy-plane is a rectangle with \(-2 \leq x \leq 2\) and \(0 \leq y \leq 2\).

Step 2: Setup Volume Integral for Estimation

To estimate the volume, we consider slicing the solid into thin slabs parallel to the yz-plane. The thickness of each slab is \(\Delta x\). In each slab, the height is given by \(z=4-x^2\) minus the \(z=0\) plane, which is the base level. The bounded length along the y-axis is 2 (from 0 to 2). Therefore, the approximate volume of each slab is \((4-x^2)(2)\Delta x\). Integrating over the range of x-values will provide a good estimate of the volume.

Step 3: Calculate the Exact Volume Using Double Integral

The exact volume can be found using a double integral. The boundaries in the xy-plane determine the limits for integration: \(x\) ranges from \(-2\) to \(2\), and \(y\) ranges from \(0\) to \(2\). The double integral becomes:\[V = \int_{0}^{2} \int_{-2}^{2} (4-x^2) \, dx \, dy\]First evaluate the inner integral with respect to \(x\):\[\int_{-2}^{2} (4-x^2) \, dx = \left[4x-\frac{x^3}{3}\right]_{-2}^{2} \]Evaluating, we find:\[= \left(8-\frac{8}{3}\right) - \left(-8+\frac{8}{3}\right) = \frac{64}{3}\]Now substitute this into the outer integral with respect to \(y\):\[\int_{0}^{2} \frac{64}{3} \, dy = \frac{64}{3}y \big|_{0}^{2} = \frac{128}{3}\]Thus, the exact volume of the solid is \(\frac{128}{3}\) cubic units.

Key Concepts

Volume of Solids

When understanding the volume of solids, especially in calculus, we aim to determine the space filled by a three-dimensional object. This task usually involves more than just basic geometry, as many solids don't fit into simple formulas. To calculate their volume, we need more advanced mathematical techniques.
Specifically, in cases where solids have curved surfaces or irregular shapes, calculus provides tools like integrals to find their volumes precisely. In this particular problem, the solid is in the first octant and is bound by both a curved surface and a flat plane. It's described by the parabolic cylinder equation, \( z = 4 - x^2 \), where the surface creates a downward opening shape. The height of the solid varies as the value of \( x \) changes.
The boundary condition \( y=2 \) helps in restricting the solid along the \( y \)-dimension, making it possible to calculate its volume within the specific region.
With these boundaries clear, we use integral calculus to find the exact volume, first estimating with thin slabs and then calculating precisely via a double integral.

Double Integrals

Double integrals are a key tool in calculus for finding areas and volumes of regions bounded by curves. They allow us to compute the volume of a solid that sits over a specific two-dimensional region in the xy-plane. This method is particularly useful because it factors in continuously varying quantities across two dimensions.
In our exercise, we use a double integral to calculate the volume of the solid bound by the surfaces \( z=4-x^{2} \) and \( y=2 \).
Here’s how the process breaks down:

• We first identify the region in the xy-plane by looking at the horizontal slice of our solid, forming a rectangle from \( x = -2 \) to \( x = 2 \) and \( y = 0 \) to \( y = 2 \).
• The height of each point in this region can be described as \( 4-x^2 \).
• We set up a double integral to sum up all these infinitesimally small volumes across this rectangle, effectively building our total volume from many tiny pieces.

First, we integrate with respect to \( x \), treating \( y \) as a constant. Then, we integrate with respect to \( y \), which gives us the total volume across the entire region.
Through this method, the exact volume, calculated as \( \frac{128}{3} \), is derived from a precise summation of these tiny slivers.

Parabolic Cylinders

Parabolic cylinders form from translating a parabola along a straight line perpendicular to its plane, resulting in a three-dimensional shape. They can be confusing at first because they do not resemble the traditional circular cylinder.
In this exercise, the cylinder is defined by the equation \( z = 4 - x^2 \). This equation describes a parabolic curve in the xz-plane that opens downwards, with its vertex at the point (0,0,4). Because it's a cylinder, this parabola extends infinitely along the y-axis.
Not like typical solids, parabolic cylinders maintain the parabolic curve shape in one plane, allowing the height to vary based on the x-coordinate, while remaining constant in terms of shape along y.

• In the given problem, the bounding plane \( y = 2 \) limits this cylinder to the section where \( 0 \leq y \leq 2 \).
• This transforms the shape into a finite solid, making calculation of properties like volume possible.

Recognizing these cylinders helps in visualizing the solid and setting up the correct integration limits, allowing for accurate volume calculation using calculus.