Problem 17 - University Calculus: Early Transcendentals

Question

Evaluate the integrals in Exercises \(7-20\). $$\int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \cos (u+v+w) d u d v d w \quad(u v w-\text { space })$$

Step-by-Step Solution

Answer

The value of the triple integral is 0.

Step 1: Understand the Problem

We need to evaluate the triple integral \( \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \cos (u+v+w) d u d v d w \). This involves integrating the function \( \cos(u+v+w) \) first with respect to \( u \), then with respect to \( v \), and finally with respect to \( w \).

Step 2: Integrate with Respect to \(u\)

Integrate \( \cos(u+v+w) \) with respect to \( u \), treating \( v \) and \( w \) as constants: \[ \int_{0}^{\pi} \cos(u+v+w) \, du = \left[ \sin(u+v+w) \right]_{0}^{\pi}. \]Evaluate the limits:\[ = \sin(\pi+v+w) - \sin(v+w). \] Since \( \sin(\pi+v+w) = -\sin(v+w) \), the result simplifies to:\[ -\sin(v+w) - \sin(v+w) = -2\sin(v+w). \]

Step 3: Integrate with Respect to \(v\)

Now integrate the resulting expression \( -2\sin(v+w) \) with respect to \( v \):\[ \int_{0}^{\pi} -2\sin(v+w) \, dv = -2 \left[ \cos(v+w) \right]_{0}^{\pi}. \] Evaluate the limits:\[ = -2(\cos(\pi+w) - \cos(w)). \]Since \( \cos(\pi+w) = -\cos(w) \):\[ = -2(-\cos(w) - \cos(w)) = 4\cos(w). \]

Step 4: Integrate with Respect to \(w\)

Finally, integrate \( 4\cos(w) \) with respect to \( w \):\[ \int_{0}^{\pi} 4\cos(w) \, dw = 4 \left[ \sin(w) \right]_{0}^{\pi}. \] Evaluate the limits:\[ = 4(\sin(\pi) - \sin(0)) = 4(0-0) = 0. \]

Step 5: Conclusion

The result of the triple integral is 0, as all terms evaluated to lead the result to zero.

Key Concepts

Multiple Integrals

Multiple integrals are a fascinating part of calculus. They allow us to calculate volumes, mass, and other properties of multidimensional systems. Think of them as an extension of single-variable integrals to higher dimensions.
In the context of the original exercise, we are dealing with a triple integral. This means we're integrating over three separate variables: \( u \), \( v \), and \( w \). Each one represents a dimension in our problem, and all of them have bounds from \( 0 \) to \( \pi \).
Calculating a triple integral generally involves nesting integrations within one another. We start with one variable, like \( u \), and proceed through others, encapsulating the final integral. By treating all other variables as constants at each step, we simplify the problem immensely.
Multiple integrals are very useful in physics and engineering, especially for calculating things like the center of mass or the moment of inertia of complex shapes. This approach helps us understand how systems behave across multiple dimensions.

Integration Techniques

When dealing with complex integrals, like our triple integral featuring three variables, techniques become vital for simplicity and effectiveness. Let's zoom in on some methods we used in the problem.
A common technique is changing the order of integration. However, in this problem, we follow the standard order. Each integration step reduces the complexity: starting with \( u \), moving to \( v \), and finally \( w \). By handling each variable separately, we break a challenging integral into more manageable pieces.
Integration by parts is another standard technique often used, although not in our specific case. It's useful when direct integration is difficult. An important part of mastering integration techniques is recognizing when simplifications occur. For example, trigonometric identities (like \( \sin(\pi + x) = -\sin(x) \) and \( \cos(\pi + x) = -\cos(x) \)) can dramatically reduce the work needed by simplifying expressions.
Finally, evaluating limits correctly is crucial, as it can distinguish between getting a correct or incorrect final answer. Mastery over a variety of techniques will make tackling integrals, like the one discussed, much easier.

Trigonometric Integrals

Trigonometric integrals often appear in calculus, involving sine, cosine, and other trigonometric functions. These can sometimes look intimidating, but with practice, they become much more approachable.
In the given integral problem, we handled a trigonometric function: \( \cos(u+v+w) \). Integrating trigonometric functions often requires knowledge of basic identities and rules to simplify terms.
As shown in the solution, while integrating \( \cos(u+v+w) \) with respect to \( u \), the function is treated like any typical cosine integral, leading to a sine function as its antiderivative. This is a standard outcome due to the differentiation relationship between sine and cosine. We used this relationship repeatedly until simplifying down to an expression only depending on \( w \).
Recognizing these relationships between trigonometric functions reduces complexity and ensures we arrive at the right answer. Additionally, understanding trigonometric identities, like the ones used here, assists in evaluating definite integrals, especially when limits such as \( \pi \) have known sine and cosine results.