Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Volume of the solid bounded by the cylinder \(y=x^{2}+2\) and the planes \(y=4, z=0\), and \(3 y-4 z=0\)

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{27\sqrt{2}}{5}\).

Step 1: Identify the Region of Integration

The solid is bounded by the cylinder described by the parabola in the plane, which is given by the equation \(y = x^2 + 2\) and is bounded by the planes \(y = 4\), \(z = 0\), and \(3y - 4z = 0\). To find the range of \(x\), set \(y = 4\) and \(y = x^2 + 2\). Solving gives \(x^2 + 2 = 4\) which results in \(x^2 = 2\), therefore, \(x = \pm \sqrt{2}\). The range of \(y\) is therefore from \(x^2 + 2\) to \(4\).

Step 2: Set Up the Integral for Volume

Considering the planes that bound the solid, \(z = 0\) is the bottom boundary, and the top boundary is given by \(z = \frac{3y}{4}\). The volume \(V\) of the solid can be calculated using the integral \[V = \int_{-\sqrt{2}}^{\sqrt{2}} \int_{x^2 + 2}^{4} \int_{0}^{\frac{3y}{4}} dz \, dy \, dx.\]

Step 3: Integrate with Respect to z

First, integrate with respect to \(z\) from 0 to \(\frac{3y}{4}\): \[\int_{0}^{\frac{3y}{4}} dz = \left[ z \right]_{0}^{\frac{3y}{4}} = \frac{3y}{4}.\]This reduces the volume integral to \[V = \int_{-\sqrt{2}}^{\sqrt{2}} \int_{x^2 + 2}^{4} \frac{3y}{4} \, dy \, dx.\]

Step 4: Integrate with Respect to y

Now integrate with respect to \(y\): \[\int (\frac{3y}{4}) \, dy = \frac{3}{8} y^2.\]Evaluating this from \(y = x^2 + 2\) to \(y = 4\) gives:\[\left[ \frac{3}{8} y^2 \right]_{x^2 + 2}^{4} = \frac{3}{8} (16) - \frac{3}{8} ((x^2+2)^2).\]Calculate this to get:\[6 - \frac{3}{8} (x^4 + 4x^2 + 4).\]

Step 5: Simplify and Integrate with Respect to x

The expression becomes:\[V = \int_{-\sqrt{2}}^{\sqrt{2}} \left( 6 - \frac{3}{8} (x^4 + 4x^2 + 4) \right) \, dx.\]Simplify further:\[= \int_{-\sqrt{2}}^{\sqrt{2}} \left( 6 - \frac{3}{8} x^4 - \frac{3}{2} x^2 - \frac{3}{2} \right) \, dx.\]\[= \int_{-\sqrt{2}}^{\sqrt{2}} \left( \frac{9}{2} - \frac{3}{8} x^4 - \frac{3}{2} x^2 \right) \, dx.\]

Step 6: Evaluate the Integral

Now integrate term-by-term:\[\int_{-\sqrt{2}}^{\sqrt{2}} \frac{9}{2} \, dx = 9\sqrt{2},\]\[\int_{-\sqrt{2}}^{\sqrt{2}} \frac{3}{2} x^2 \, dx = \frac{3}{2} \cdot 2 = 3\sqrt{2},\]\[\int_{-\sqrt{2}}^{\sqrt{2}} \frac{3}{8} x^4 \, dx = \frac{3}{8} \cdot \frac{16\sqrt{2}}{5} = \frac{3}{5}\sqrt{2}.\]Add these results together:\[9\sqrt{2} - 3\sqrt{2} - \frac{3}{5}\sqrt{2} = 6\sqrt{2} - \frac{3}{5}\sqrt{2}.\]Simplify to find the volume.The final calculation simplifies as: 6\sqrt{2} - \frac{3}{5}\sqrt{2} = \frac{27\sqrt{2}}{5}.

Key Concepts

Volume of Solids

Volume is an essential concept in multivariable calculus, especially when calculating the three-dimensional space occupied by a solid object. Here, we're looking at a solid bounded by a cylindrical surface and certain planes.
The goal is to determine how much space this solid occupies, or its volume. By using calculus, we can evaluate this volume using definite integrals, integrating over the entire region the solid extends.
We need to break the problem down into manageable parts—like slices of bread in a loaf—to effectively compute the volume. By integrating the slices, which are determined by the boundaries given by equations, we calculate the total volume occupied by the solid.

• Understanding the boundaries is crucial. In this problem, those boundaries are provided by the cylinder and specific planes (surfaces), which prevent the solid from extending infinitely.
• To effectively set up our integration, we need to determine the limits on each axis (here, we focus on x, y, and z).

Integration Techniques

Integration is a fundamental technique in calculus that involves finding the whole by summing its parts. In multivariable calculus, it's used in a more advanced way to calculate things like volume.
Here, the method used is triple integration, which means integrating with respect to three different variables, one after the other.
We start with the innermost integral and move outward. The key here is setting up the integrals correctly based on the description of the volume's boundaries.

• First, integrate with respect to the innermost variable (in this case, z) to calculate the height at a point (x, y).
• Then, integrate with respect to y to account for changes along the cylinder's height.
• Finally, integrate with respect to x to incorporate variations across the base of the solid.

This systematic approach ensures that all parts of the solid's volume are covered, providing an accurate result.

Cylindrical Coordinates

Cylindrical coordinates are a three-dimensional extension of the two-dimensional polar coordinates. These coordinates are particularly useful in problems involving symmetry around an axis, like the solid bounded by a cylindrical surface.
In this exercise, while it was easier to work with Cartesian coordinates initially, converting to cylindrical coordinates could simplify the process for some solids.
We use cylindrical coordinates when it helps to describe the region more naturally, especially when dealing with circular or cylindrical shapes. The coordinates consist of:

• Radial distance (r) from the axis.
• Angular coordinate (θ), similar to the angle in polar coordinates.
• Height (z), the same as in Cartesian coordinates.

These coordinates highlight the symmetry in problems involving cylinders or rotationally symmetric shapes, which can simplify the integration process by aligning the integrals with the geometry of the shape.

Definite Integrals

A definite integral calculates the accumulation of quantities, like area or volume, between defined limits. In the context of multivariable calculus, definite integrals can be extended to measure three-dimensional space.
We use definite integrals to precisely calculate the volume of a solid by summing the 'slices' under specific conditions, dictated by lower and upper limits.
In this problem, definite integrals are used to ensure the volume calculation is bound by the specified limits, such as the starting and ending points of the solid along each axis.

• The limits of integration for each variable (x, y, z) are essential in defining the region of the solid we analyze.
• These limits are derived from the equations representing the boundaries of the solid, ensuring we integrate over the correct region.

By applying definite integrals, we obtain a precise measure of the volume, taking into account the full scope of its bounding surfaces and planes.