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Chapter 5: Problem 15

Evaluate the integrals in Exercises 15 to 23. $$\int_{0}^{1} \int_{1}^{2} \int_{2}^{3} \cos [\pi(x+y+z)] d x d y d z$$

Short Answer

Expert verified
The integral evaluates to 0.

Step by step solution

01

Understanding the Triple Integral

We are given the integral \(\int_{0}^{1} \int_{1}^{2} \int_{2}^{3} \cos [\pi(x+y+z)] \, d x \, d y \, d z\).This indicates we will evaluate the integral over three variables: \(x\), \(y\), and \(z\). The integral will involve evaluating the innermost part first (\(dx\)), then \(dy\), and finally \(dz\).
02

Evaluate the Integral with Respect to x

We first evaluate \(\int_{2}^{3} \cos[\pi(x+y+z)] \, dx\). Here, \(y\) and \(z\) are constants. Let \(u = \pi(x+y+z)\). Then \(du = \pi \, dx\) so \(dx = \frac{du}{\pi}\). The limits for \(x\) are from 2 to 3, hence for \(u\), they are \(\pi(y+z+2)\) and \(\pi(y+z+3)\).
03

Integrate with Substitution

After substituting, we get \(\int_{\pi(y+z+2)}^{\pi(y+z+3)} \cos(u) \, \frac{du}{\pi}\) = \frac{1}{\pi} [\sin(u)]_{\pi(y+z+2)}^{\pi(y+z+3)}). Calculating this integral, we find: \[\frac{1}{\pi} \left(\sin(\pi(y+z+3)) - \sin(\pi(y+z+2))\right)\].Since \(\sin\) of any multiple of \(\pi\) is zero, this integral evaluates to 0.
04

Integrate with Respect to y and z

Since the innermost integral evaluates to zero, there's no need to integrate with respect to \(y\) and \(z\) as further integration will still yield zero. Hence\(\int_{1}^{2}\int_{0}^{1} 0 \, dy \, dz = 0\).

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Key Concepts

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

Integration by Substitution
Integration by substitution is a powerful technique often used to simplify complex integrals. It works by changing the variable of integration to transform the integral into a new, often simpler form.
In our problem, \(\int_{0}^{1} \int_{1}^{2} \int_{2}^{3} \cos [\pi(x+y+z)] \, dx \, dy \, dz\), the innermost integral is initially evaluated by letting \(u = \pi(x+y+z)\).
This substitution helps isolate one variable from the others, essentially converting the problem into evaluating the cosine integral of \(u\).
  • The derivative \(du = \pi \, dx\) rearranges to \(dx = \frac{du}{\pi}\), simplifying the integral.
  • Upon substitution, the limits of integration need to be adjusted accordingly, converting from \(x\) to \(u\).
Replacing variables and changing the limits make it easier to evaluate trigonometric integrals, demonstrating the effectiveness of integration by substitution.
Cosine Function
The cosine function, \(\cos(u)\), is a key player in this integral. It defines the wave's behavior over cycles represented by multiples of \(\pi\).
Cosine is periodic and even, meaning \(\cos(-u) = \cos(u)\), which simplifies calculating integrals involving these functions.
In our case, the integral involved substituting the expression \(\cos[\pi(x+y+z)]\), which we reduced to \(\cos(u)\).
  • The integral's value is often related to its periodic nature 鈥� since \(\sin(\pi n) = 0\) for any integer \(n\), integrals resolving to this form tend to zero.
  • The ultimate goal of evaluating \(\cos(u)\) over its limits demonstrated the property \(\sin(\pi(y+z+3)) - \sin(\pi(y+z+2)) = 0\).
Understanding these properties of the cosine function helps in efficiently resolving integrals and predicting outcomes, especially when in the context of waveforms or oscillatory motion.
Nested Integrals
Nested integrals, or repeated integrals, involve integrating a function multiple times over different variables. This integral \(\int_{0}^{1} \int_{1}^{2} \int_{2}^{3} \cos [\pi(x+y+z)] \, dx \, dy \, dz\) is a triple integral, indicating it involves three layers.
The process is hierarchical:
  • Start from the innermost integral and progressively move outward.
  • In this exercise, the first integral concerning \(x\) completely determined the outcome - once evaluated to zero, the subsequent evaluations for \(y\) and \(z\) inherently remain zero.
Each layer of integration considers external variables as constants, narrowing the focus to just one at a time. This structured approach makes it feasible to solve integrals encountered in multi-dimensional problems across physics, engineering, and mathematics.

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Most popular questions from this chapter

(a) Sketch the region for the integral \(\int_{0}^{1} \int_{0}^{x} \int_{0}^{y} f(x, y, z) d z d y d x.\) (b) Write the integral with the integration order \(d x d y d z.\)

Evaluate the following iterated integrals: \(\begin{array}{ll}\text { (a) } \int_{0}^{1} \int_{0}^{1}\left(1-x^{3}+x y\right) d x d y & \text { (c) } \int_{1}^{2} \int_{2}^{4}\left(\frac{x}{y}+\frac{y}{x}\right) d x d y \\\ \text { (b) } \int_{0}^{\pi / 2} \int_{-\pi / 2}^{\pi / 2} \cos x \sin y d x d y & \text { (d) } \int_{0}^{\pi / 4} \int_{0}^{\pi / 4} \tan x \sec ^{2} y d x d y\end{array}\)

Describe the given region as an elementary region. The region between the cone \(z=\sqrt{x^{2}+y^{2}}\) and the paraboloid \(z=x^{2}+y^{2}.\)

Let \(f:[0,1] \times[0,1] \rightarrow \mathbb{R}\) be defined by $$ f(x, y)=\left\\{\begin{array}{ll} 1 & x \text { rational } \\ 2 y & x \text { irrational. } \end{array}\right. $$ Show that the iterated integral \(\int_{0}^{1}\left[\int_{0}^{1} f(x, y) d y\right] d x\) exists but that \(f\) is not integrable.

Describe the given region as an elementary region. The region inside the sphere \(x^{2}+y^{2}+z^{2}=1\) and above the plane \(z=0.\)
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