Problem 19 - APEX Calculus

Question

Evaluate the triple integral. $$ \int_{0}^{\pi} \int_{0}^{1} \int_{0}^{z}(\sin (y z)) d x d y d z $$

Step-by-Step Solution

Answer

The triple integral evaluates to \(\pi\).

Step 1: Understand the Order of Integration

The given integral is a triple integral in the order \(dx\), \(dy\), \(dz\). Begin with integrating with respect to \(x\), followed by \(y\), and finally \(z\). Each integration removes one of the integral signs.

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

The function to integrate is \( \sin(yz) \). Since \( \sin(yz) \) is constant with respect to \(x\), we treat \( \sin(yz) \) as a constant while integrating over \(x\):\[\int_{0}^{z} \sin(yz) \, dx = \sin(yz) \cdot [x]_{0}^{z} = \sin(yz) \cdot (z - 0) = z \sin(yz).\]

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

Now, integrate the resulting expression \(z \sin(yz)\) with respect to \(y\):\[\int_{0}^{1} z \sin(yz) \, dy.\]Substitute \(u = yz\), which gives \(du = z \, dy\) or \(dy = \frac{du}{z}\). Adjust the limits for \(y = 0\) to \(y = 1\):- When \(y = 0\), \(u = 0\).- When \(y = 1\), \(u = z\).\[z \int_{0}^{z} \sin(u) \cdot \frac{du}{z} = \int_{0}^{z} \sin(u) \, du.\]The integral of \( \sin(u)\) is \(-\cos(u)\), thus:\[[-\cos(u)]_{0}^{z} = -\cos(z) - (-\cos(0)) = -\cos(z) + 1.\]

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

Integrate the result \(-\cos(z) + 1\) with respect to \(z\):\[\int_{0}^{\pi} (-\cos(z) + 1) \, dz.\]This splits into two separate integrals:1. \(\int_{0}^{\pi} -\cos(z) \, dz = -\sin(z) \bigg|_{0}^{\pi} = -\sin(\pi) + \sin(0) = 0\).2. \(\int_{0}^{\pi} 1 \, dz = [z]_{0}^{\pi} = \pi - 0 = \pi\).Thus, the combined result is \(0 + \pi\).

Step 5: Combine the Results

After performing all integrations, the final result is simply the outcome from the last step: \(\pi\). This is the value of the original triple integral.

Key Concepts

Understanding Order of Integration

In a triple integral, the order of integration refers to the sequence in which we integrate the variables. This typically occurs from the innermost integral to the outermost. In the given problem, we are dealing with the order \(dx\), \(dy\), \(dz\).
Begin by integrating with respect to \(x\), followed by \(y\), and finally \(z\). Each completed integration reduces the complexity of the integral, removing one layer at a time.
Choosing the correct order is important, as it can simplify computations and help ensure the bounds fit together nicely.
In this problem, the order of \(dx\), \(dy\), and \(dz\) was appropriate because each variable had a clear range dependent on the preceding variable.

Substitution Method in Integration

The substitution method is an essential technique in calculus that simplifies integrals by transforming them into a more manageable form. Essentially, it involves substituting a part of the integral with a new variable to make integration easier.
In the second integration step of the problem, for integrating \(z \sin(yz)\) with respect to \(y\), we set \(u = yz\). This substitution changes the variables, allowing us to convert a complex integral into a more straightforward one.

• Substitute \(u = yz\), resulting in \(du = z \, dy\).
• The differential \(dy\) is replaced by \(\frac{du}{z}\), integrating with respect to \(u\).
• Convert the limits: as \(y\) changes from 0 to 1, \(u\) changes from 0 to \(z\).

This simplifies the integral to \(\int_{0}^{z} \sin(u) \, du\). Substitution transformed a difficult part of the problem into a recognizable and solvable format.

Using Integration by Parts

Integration by parts is a useful technique when dealing with products of functions. While it wasn't explicitly required in this particular problem, understanding it is crucial for similar integration challenges.
The method is based on the product rule for differentiation, but applied in reverse for integration. The formula is:
\[\int u \, dv = uv - \int v \, du\]
where \(u\) and \(dv\) are chosen parts of the integrand.
It's often used when substitution isn't sufficient, typically with polynomial and exponential functions or products involving trigonometric functions. Choosing which parts to assign as \(u\) and \(dv\) can greatly affect the outcome.

• Pick \(u\) as the function that becomes simpler when differentiated.
• Choose \(dv\) such that its antiderivative is easily recognizable.

Integration by parts can be particularly helpful when dealing with repeated patterns such as polynomials multiplied by trig functions.

Sine Function Integration

Integrating trigonometric functions like sine is a fundamental skill in calculus. When dealing with \(\sin(u)\), its integral is directly related to the cosine function.
The integral of \(\sin(u)\) is \(-\cos(u)\).
This was applied in the problem after using the substitution method:

• Firstly, simplify the integral to \(\int_{0}^{z} \sin(u) \, du\).
• Integrate to get \(-\cos(u)\).
• Easily evaluate the definite integral as \([-\cos(z) + 1]\).

Understanding these standard integrals helps in quickly solving more complex problems. Trigonometric integrals often reappear in various contexts, so becoming comfortable with their patterns is very beneficial.