Chapter 5: Problem 42

Find the volume between the planes \(z=2 x+3 y+6\) and \(z=2 x+7 y+8\) and over the square in the \((x, y)\) plane with vertices \((0,0),(1,0),(0,1),(1,1)\).

Short Answer

Expert verified

The volume is 4 cubic units.

Step by step solution

Understand the assignment

The goal is to find the volume between the planes given by the equations: 1. Plane 1: \( z = 2x + 3y + 6 \) 2. Plane 2: \( z = 2x + 7y + 8 \) over the square in the \( (x, y) \) plane with vertices: \( (0,0) \), \( (1,0) \), \( (0,1) \), \( (1,1) \).

Set up the volume integral

The volume formula between two surfaces is given by the double integral of the difference of the functions describing the surfaces over the specified region. We will integrate: \( V = \int_{0}^{1} \int_{0}^{1} [(2x + 7y + 8) - (2x + 3y + 6)] \, dy \, dx \)

Simplify the integrand

First, simplify the integrand: \( (2x + 7y + 8) - (2x + 3y + 6) = 4y + 2 \). Therefore, the integral becomes: \( V = \int_{0}^{1} \int_{0}^{1} (4y + 2) \, dy \, dx \)

Integrate with respect to y

Integrate the function \( 4y + 2 \) with respect to \( y \): \( \int_{0}^{1} (4y + 2) \, dy = [2y^2 + 2y]_{0}^{1} = (2(1)^2 + 2(1)) - (2(0)^2 + 2(0)) = 2 + 2 = 4 \)

Integrate with respect to x

Integrate the result with respect to \( x \): \( \int_{0}^{1} 4 \, dx = 4x \bigg|_{0}^{1} = 4(1) - 4(0) = 4 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Double Integral

A double integral allows us to compute the volume under a surface over a given region in the xy-plane. It's an extension of the single integral, where instead of integrating over a line, we integrate over an area. In mathematical terms, a double integral over a region R can be written as: \[\text{V} = \int \int_R f(x, y) \ dx \ dy \] In the given exercise, the function f(x, y) is the difference between the two planes. Hence, using the double integral, we calculate the cumulative volume slice by slice in the given region.

Volume Calculation

To calculate the volume between two planes, we use the double integral of the difference of their equations over the defined region. This process can be simplified into smaller steps:

Determine the equations of the planes involved.
Identify the region over which to integrate.
Set up the double integral of the difference between the plane equations.
Simplify the integrand.
Evaluate the inner integral first, followed by the outer integral.

In our problem, the equations of the planes are given by z = 2x + 3y + 6 and z = 2x + 7y + 8. The region is a square with vertices (0,0), (1,0), (0,1), and (1,1). Our simplified integrand, 4y + 2, is what we need to integrate over this region.

Multivariable Calculus

Multivariable calculus extends calculus concepts to functions of multiple variables. This includes finding volumes under surfaces, optimization problems, and surface integrals.
In multivariable calculus, integration can be applied to functions of two or more variables, allowing for the computation of areas, volumes, and other properties within a multidimensional space.
In the given exercise, we observe:

The boundaries of the region of integration are clearly defined in the xy-plane.
The double integral accounts for variations in both x and y dimensions.
Integral computations often split into simpler, sequential single-variable integrals.

Understanding multivariable calculus is crucial for tackling advanced problems in physics, engineering, and other sciences, where phenomena depend on multiple variables simultaneously.