Problem 20 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

\( \frac{16}{3} \ln 8 \)

Step 1: Understand the Integral Setup

This is a triple integral in pqr-space. We need to evaluate \( \int_{0}^{7} \int_{0}^{2} \int_{0}^{\sqrt{4-q^{2}}} \frac{q}{r+1} dp dq dr \). This means the integral is evaluated first with respect to \( p \), then \( q \), and finally \( r \).

Step 2: Integrate with Respect to p

Perform the integration of \( \frac{q}{r+1} \) with respect to \( p \) from \( 0 \) to \( \sqrt{4-q^2} \). This results in:\[\int_{0}^{\sqrt{4-q^{2}}} \frac{q}{r+1} dp = \frac{q}{r+1} \cdot p \Bigg|_{0}^{\sqrt{4-q^{2}}} = \frac{q \cdot \sqrt{4-q^{2}}}{r+1}.\]

Step 3: Integrate with Respect to q

Now integrate the result \( \frac{q \sqrt{4-q^{2}}}{r+1} \) with respect to \( q \) from \( 0 \) to \( 2 \):\[\int_{0}^{2} \frac{q \sqrt{4-q^{2}}}{r+1} dq.\]To solve this, consider a substitution like \( u = 4 - q^2 \), \( du = -2q dq \), changing the limits and simplifying appropriately.

Step 4: Change of Variables for Integration

For \( u = 4 - q^2 \), when \( q = 0, u = 4 \) and when \( q = 2, u = 0 \). Rewriting, we have:\[-\frac{1}{2(r+1)} \int_{4}^{0} \sqrt{u} \, du = \frac{1}{2(r+1)} \int_{0}^{4} \sqrt{u} \, du.\]

Step 5: Evaluate Integral with Respect to u

Evaluate the integral \( \int_{0}^{4} \sqrt{u} \, du \):\[= \int_{0}^{4} u^{1/2} \, du = \left[ \frac{2}{3} u^{3/2} \right]_{0}^{4} = \frac{2}{3} \cdot (4^{3/2} - 0^{3/2}) = \frac{2}{3} \cdot 8 = \frac{16}{3}.\]

Step 6: Integrate with Respect to r

Now integrate the result from the previous integral with respect to \( r \) from \( 0 \) to \( 7 \):\[\int_{0}^{7} \frac{16}{3(r+1)} dr.\]This results in:\[\frac{16}{3} \int_{0}^{7} \frac{1}{r+1} dr = \frac{16}{3} \ln |r+1| \Big|_{0}^{7} = \frac{16}{3} (\ln 8 - \ln 1) = \frac{16}{3} \ln 8.\]

Step 7: Final Result

The final evaluated integral is \( \frac{16}{3} \ln 8 \).

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus used to simplify integrals, especially when dealing with complex expressions. This method is akin to reversing the chain rule from differentiation.
In our triple integral example, substitution is particularly handy for the integral with respect to \( q \). Specifically, by setting \( u = 4 - q^2 \), we transform the integral into a more manageable form.

• Step 1 is to identify a substitution that simplifies the integral. In this case, \( u = 4 - q^2 \) was chosen to simplify the \/term \( \sqrt{4-q^2} \).
• Calculate \( du \). For \( u = 4 - q^2 \), we have \( du = -2q \, dq \).
• Adjust the limits of integration to match the new variable \( u \). When \( q = 0, u = 4 \) and when \( q = 2, u = 0 \).

This substitution turns the complex integral into a standard form, making it easier to integrate.

Change of Variables

The change of variables technique is a key aspect of multivariable calculus. It allows us to transform an integral into a different coordinate system or variable set that is easier to handle.
In the given triple integral, after substituting \( u \) for \( q \), we effectively change the integral from \( q \)-terms into \( u \)-terms.

• This approach often involves transforming the integral's limits, as seen when \( q \) changes into \( u \). For instance, \( q \)'s limit of 0 translates to a \( u \) limit of 4.
• Additionally, a negative sign emerged due to \( du = -2q \, dq \), which flipped the integration limits.
• Rewriting the integral involves addressing these sign changes and solving from 0 to 4 instead of 4 to 0.

By carefully managing these transformations, the computation of more intricate integrals becomes feasible through more straightforward expressions.

Integration Techniques

Integration techniques are crucial tools for solving integrals of varying complexity. For triple integrals, understanding these techniques greatly improves efficiency.
Our example demonstrates several aspects:

• **Integration with respect to each variable**: The process involves integrating step-by-step, starting with \( p \), then \( q \), finally \( r \), allowing us to simplify integrals one variable at a time.
• **Transformations**: Using substitutions like \( u = 4 - q^2 \) to simplify the expression before proceeding with integration.
• **Final integration result**: Finding the result after integrating with respect to all variables gives us \( \frac{16}{3} \ln 8 \).

These techniques ensure accurate and effective evaluation of even the most complex integrals by breaking them into simpler parts.