Problem 20 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(7-20\). $$ \int_{0}^{7} \int_{0}^{2} \int_{0}^{\sqrt{4-q^{2}}} \frac{q}{r+1} d p d q d r \quad(p q r-\text { space }) $$

Step-by-Step Solution

Answer

\( \frac{8}{3} \ln(8) \)

Step 1: Inner Integral with Respect to p

Let's begin by evaluating the innermost integral, which is with respect to the variable \( p \). The integral given is \( \int_{0}^{\sqrt{4-q^{2}}} \frac{q}{r+1} \, dp \). Since the integrand \( \frac{q}{r+1} \) is constant with respect to \( p \), we can easily integrate it over \( p \). Thus, the integral becomes:\[\int_{0}^{\sqrt{4-q^{2}}} \frac{q}{r+1} \, dp = \frac{q}{r+1} \times p \bigg|_{0}^{\sqrt{4-q^{2}}}\]Evaluating the integral, we get:\[\frac{q}{r+1} \times \sqrt{4-q^{2}}\]

Step 2: Second Integral with Respect to q

Next, substitute the result from the previous step into the second integral, which is with respect to \( q \). The integral is now:\[\int_{0}^{2} \frac{q \sqrt{4-q^{2}}}{r+1} \, dq\]To evaluate this, we can use substitution. Let \( u = 4 - q^2 \), then \( du = -2q \, dq \), or \( -\frac{1}{2} \, du = q \, dq \). Also, change the limits of integration for \( q = 0 \) to \( q = 2 \), which gives \( u = 4 \) to \( u = 0 \). The integral becomes:\[-\frac{1}{2(r+1)} \int_{4}^{0} \sqrt{u} \, du = \frac{1}{2(r+1)} \int_{0}^{4} \sqrt{u} \, du\]This evaluates to:\[\frac{1}{2(r+1)} \times \frac{2}{3} u^{3/2} \bigg|_{0}^{4} = \frac{1}{3(r+1)} \times (8) = \frac{8}{3(r+1)}\]

Step 3: Outer Integral with Respect to r

Finally, we need to integrate with respect to the variable \( r \) from the result of the previous steps. The integral now becomes:\[\int_{0}^{7} \frac{8}{3(r+1)} \, dr\]This is a straightforward integral to evaluate:Let \( u = r+1 \), then \( du = dr \). The limits change from \( r = 0 \) to \( r = 7 \), hence \( u = 1 \) to \( u = 8 \). The integral becomes:\[\frac{8}{3} \int_{1}^{8} \frac{1}{u} \, du\]which is:\[\frac{8}{3} \ln|u| \bigg|_{1}^{8} = \frac{8}{3} (\ln 8 - \ln 1) = \frac{8}{3} \ln 8\]

Step 4: Final Calculation

Since \( \ln 1 = 0 \), the final evaluation of the integral becomes:\[\frac{8}{3} \ln(8)\]

Step 5: Conclusion

Thus, the evaluated integral is \( \frac{8}{3} \ln(8) \).

Key Concepts

Iterated Integrals

Triple integrals expand on single and double integrals by integrating a function of three variables. When performing these integrations, we often break the problem down into a series of integrals known as **iterated integrals**. This allows us to evaluate the integral in steps, focusing on one variable at a time.
In the given exercise, the integration is performed first with respect to \( p \), then \( q \), and finally \( r \). This order is crucial for correctly evaluating the integral.

• For the **innermost integral** \( \int_{0}^{\sqrt{4-q^{2}}} \frac{q}{r+1} \, dp \), the expression \( \frac{q}{r+1} \) is constant with respect to \( p \), simplifying our work significantly.
• The **next layer** of integration \( \int_{0}^{2} \frac{q \sqrt{4-q^{2}}}{r+1} \, dq \) demands a bit more finesse. Often we use substitution techniques to make the integration smoother.
• The **outermost** layer \( \int_{0}^{7} \frac{8}{3(r+1)} \, dr \) is just as crucial, as each preceding step provides the groundwork for it.

By approaching each integration systematically, iterated integrals transform complex problems into manageable tasks. This approach is key in cases like this, streamlining the computation process.

Polar Coordinates

The concept of polar coordinates often comes in helpful situations, especially for integrals involving circular or spherical boundaries. Though the given exercise doesn't explicitly translate to polar coordinates, it's vital to recognize where they may apply because of the circular constraints in the limits.
For example, the term \( \sqrt{4-q^2} \) may hint at a circular boundary, reminiscent of a circle with radius 2. This suggests that some parts of the integral might naturally align with polar or cylindrical coordinates.
Polar coordinates replace \( (x, y) \) with \( (r, \theta) \):

• **\( r \)** represents the radius or distance from the origin.
• **\( \theta \)** stands for the angle from the positive x-axis.

The idea is that whenever you see square roots like \( \sqrt{A^2 - B^2} \), considering polar coordinate substitution can be useful. However, in our triple integral, we are instead focusing on substitution for integration ease, as the problem suggests a different structure than a direct polar transformation.

Substitution Method

The substitution method is a popular technique in calculus used to simplify integrals. In the context of our problem, we utilize substitution to handle the integral with respect to \( q \), specifically in \( \int_{0}^{2} \frac{q \sqrt{4-q^{2}}}{r+1} \, dq \).
The substitution technique involves finding a new variable \( u \) which transforms the integral into a simpler form. In our exercise, we set \( u = 4 - q^2 \), leading to \( du = -2q \, dq \). Consequently, the integral changes from variable \( q \) to \( u \), with the limits transformed accordingly.

• **Benefit**: This transformation simplifies integrals involving square roots and polynomial expressions.
• **Change of Limits**: When performing substitution, always ensure the integration limits change to match the new variable \( u \).
• **Solving**: The integral simplifies to \(-\frac{1}{2(r+1)} \int_{4}^{0} \sqrt{u} \, du \).

Substitution is a powerful tool, allowing complex integrals to become manageable. It gives us clarity and simplifies the problem, transforming it into one we can easily solve using basic integration techniques.