Question

Evaluate the triple integral. $$ \int_{-\pi / 2}^{\pi / 2} \int_{0}^{\pi} \int_{0}^{\pi}(\cos x \sin y \sin z) d z d y d x $$

Short answer

The value of the triple integral is 8.

Step-by-step solution

Understand the Integral Bounds

We are asked to evaluate the triple integral \(\int_{-\pi / 2}^{\pi / 2}\int_{0}^{\pi}\int_{0}^{\pi}(\cos x \sin y \sin z)\ dz\ dy\ dx\). This integral first integrates the function with respect to \(z\), then \(y\), and finally \(x\).

Integrate with respect to z

Start by integrating \(\cos x \sin y \sin z\) with respect to \(z\). The antiderivative of \(\sin z\) is \(-\cos z\). So the integral becomes:\[\int_{0}^{\pi} \sin z\ dz = [-\cos z]_{0}^{\pi} = -\cos(\pi) + \cos(0) = 1 + 1 = 2.\]

Substitute First Integral Result

Substitute the result of the first integral into the next integral:\[\int_{0}^{\pi} \int_{-rac{\pi}{2}}^{\frac{\pi}{2}} \cos x \sin y \cdot 2 \ dy\ dx.\]This simplifies to:\[2 \int_{0}^{\pi} \sin y \ dy \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\ dx.\]

Integrate with respect to y

Now integrate \(2\sin y\) with respect to \(y\):\[2 \int_{0}^{\pi} \sin y\ dy = 2 [-\cos y]_{0}^{\pi} = 2 (-\cos(\pi) + \cos(0)) = 2 (1 + 1) = 4.\]

Substitute Second Integral Result

Substitute the result of the second integral into the next integral:\[4 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\ dx.\]

Integrate with respect to x

Finally, integrate \(4\cos x\) with respect to \(x\):\[4 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\ dx = 4 [\sin x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 4 (\sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})) = 4(1 - (-1)) = 4 \times 2 = 8.\]

Conclude the Evaluation

The result of the triple integral is 8. Each step involved evaluating an integral and substituting it back into the remaining expression until the final integral was evaluated.

Key concepts

Integration by Parts

Integration by parts is a technique often used to evaluate integrals where the standard methods of integration are not easily applicable. This method is derived from the product rule for differentiation and helps break down an integral into more manageable parts. The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]Here, the choice of \(u\) and \(dv\) is crucial, and it significantly depends on the function at hand. One useful strategy is the ILATE rule, which helps prioritize the choice of \(u\) based on the type of function present.Integration by parts is particularly helpful in dealing with products of polynomial, exponential, or trigonometric functions. It is not necessary for all triple integrals, but knowing it broadens your ability to tackle more complex integrations that may arise. In the context of our exercise, integration by parts wasn't needed due to the straightforward nature of the functions, but it’s a powerful tool to have when dealing with more complicated integrals.

Iterated Integrals

Iterated integrals are double, triple, or higher-dimensional integrals where we integrate a function over multiple variables, one at a time. In our exercise, we evaluated a triple integral, which meant integrating over three different variables: first \(z\), then \(y\), and finally \(x\).The process involves performing each integration sequentially. In each step:

• We view the remaining variables as constants while integrating with respect to one variable.
• Once an inner integral is evaluated, its result is used for the next outer integral.
• This continues until all integrals have been evaluated.

This method of evaluating step-by-step integration is essential for functions in multiple dimensions. The bounds of each integral limit the extent of each variable, which is crucial to correctly perform the integration. Iterated integrals are fundamental in computing volumes, areas, and other quantities in higher dimensions.

Antiderivative

An antiderivative of a function is another function whose derivative is the original function. Finding an antiderivative is central to the process of integration, helping us move backward from derivatives to original functions. When integrating, identifying the correct antiderivative allows us to evaluate the integral over a specified range by substituting the boundary values.In our exercise, determining the antiderivative of \(\sin z\) involved identifying \(-\cos z\) as the correct form, due to the relation \(\frac{d}{dz}(-\cos z) = \sin z\). Similarly, understanding the antiderivatives of other trigonometric functions is vital:

• \(\int \cos(x) \, dx = \sin(x) + C\)
• \(\int \sin(x) \, dx = -\cos(x) + C\)

Grasping these basic antiderivatives allows students to effectively solve integrals within and beyond textbook exercises. Therefore, mastering this concept not only simplifies problems but also aids in understanding the broader applications of calculus.