Problem 17 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The value of the integral is 0.

Step 1: Identify the Integral

The exercise requires evaluating a triple integral over the cube defined by \(0 \leq u, v, w \leq \pi\). The integrand is \(\cos(u+v+w)\).

Step 2: Simplify Using Symmetry

Notice that \(\cos(u+v+w)\) is an odd function around \((u,v,w)=(\frac{\pi}{2},\frac{\pi}{2},\frac{\pi}{2})\). The full integration region from 0 to \(\pi\) in each variable suggests potential symmetry; specifically, over half of the region, the function will cancel out to zero due to being centered at points that yield equal positive and negative values.

Step 3: Evaluate the Inner Integral

Consider the first integration with respect to \(u\), keeping \(v\) and \(w\) constant: \[\int_0^\pi \cos(u+v+w) \, du = \sin(u+v+w) \bigg|_0^\pi.\]This evaluates to zero because \(\sin(\pi+x) = -\sin(x)\) and \(\sin(0+x) = \sin(x)\). Thus, the expression becomes \(\sin(\pi) - \sin(v+w) = 0 - \sin(v+w)\), leading to a whole setup that is symmetric and zero.

Step 4: Extend Integration Over All Variables

Applying the same logic, integrate over \(v\): \[\int_0^\pi 0 \, dv = 0.\]And then over \(w\): \[\int_0^\pi 0 \, dw = 0.\]Thus, the whole triple integral simplifies directly given symmetry causing cancellation.

Step 5: Final Evaluation and Conclusion

Since the integrand \(\cos(u+v+w)\) is symmetric and results in cancellation due to its symmetry about the midpoint of the limits, the final value of the whole triple integral is zero.

Key Concepts

Symmetry in Calculus

Symmetry plays a critical role in simplifying complex integrals in calculus. In our exercise, the integral we are considering is taken over a cubic region with equal limits for each variable, from 0 to \( \pi \).
This setup provides the perfect conditions to analyze symmetry. When dissecting symmetry, note that the function \( \cos(u+v+w) \) is odd around the midpoint \((\frac{\pi}{2}, \frac{\pi}{2}, \frac{\pi}{2})\).
Recognizing this odd symmetry is key: it suggests that for any input values in the integral's region, there are corresponding values that will cancel each other out.
This is because the cosine function will flip from positive to negative equally across the central point of integration.
This symmetry results in the contribution to the integral from opposing points being zero. By identifying this, we save ourselves a significant amount of work, as we don't need to calculate each part by hand when the result is immediately evident as zero due to this symmetrical behavior.

Integration Techniques

In multivariable calculus, tackling triple integrals demands efficient integration techniques. One approach is assessing the order of integration.
For the problem at hand, evaluating over \( u \), then \( v \), and finally \( w \), flags an underlying simplicity. The traditional integration requires handling each nested integral independently.Here's why this order matters:

• The integrand \( \cos(u+v+w) \) simplifies beautifully when isolated for each variable.
• Computing the integral, say with respect to \( u \), allows us to identify \/ react to any symmetric tendencies.

The advantage becomes evident when, after integrating over \( u \), you reveal zero due to periodic properties of sine. Repeating this pattern for \( v \) and \( w \) confirms the integrity of selecting our integral bounds.
Employing such techniques masterfully demystifies what initially seems intricate: though the complexity rises with dimensions, symmetry continues to be a saving grace.

Multivariable Calculus

Navigating multivariable calculus introduces the challenge of visualizing and manipulating functions within higher dimensions. A triple integral covers three dimensions, crucially extending beyond familiar single-variable boundaries.
Here, you calculate over a volume rather than a mere line or surface.To approach such problems:

• Visualize the region of integration, which in this problem is a cube with edges from 0 to \( \pi \).
• Designate bounds explicitly to see how each variable influences the outcome.

In this exercise, the variables \( u \), \( v \), and \( w \) define points in a 3D space and the function \( \cos(u+v+w) \) acts within this space as a three-dimensional wave.
Grasping how each variable interacts necessitates a blend of geometric intuition and algebraic manipulation, especially when considering the impact of each variable iteratively.
As multivariable calculus grows in complexity, integrating skills from one-dimensional calculus, staying attuned to symmetry, and leveraging spatial reasoning stand as invaluable tools.