Problem 22 - University Calculus: Early Transcendentals

Question

Evaluate the spherical coordinate integrals. $$\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{2}(\rho \cos \phi) \rho^{2} \sin \phi d \rho d \phi d \theta$$

Step-by-Step Solution

Answer

The integral evaluates to \(8\pi\).

Step 1: Understand the Integral Setup

The given triple integral \[ \int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{2}(\rho \cos \phi) \rho^{2} \sin \phi \, d \rho \, d \phi \, d \theta \]is set up in spherical coordinates. Here, \(\rho\) is the radial distance, \(\phi\) is the polar angle, and \(\theta\) is the azimuthal angle. The limits indicate integration over a spherical cap.

Step 2: Simplify the Integral

Rewrite the integrand as \((\rho \cos \phi) \rho^{2} \sin \phi = \rho^{3} \cos \phi \sin \phi.\)This step simplifies the expression to make integration easier.

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

Focus on the innermost integral:\[\int_{0}^{2} \rho^{3} \cos \phi \sin \phi \, d \rho.\]This is a straightforward power rule integration:\[\frac{\rho^{4}}{4} \Bigg|_{0}^{2} = \frac{2^4}{4} - 0 = 4.\]Thus, the result is \(4 \cos \phi \sin \phi\).

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

Simplify to the middle integral:\[\int_{0}^{\pi/4} 4 \cos \phi \sin \phi \, d \phi.\]Use the identity \(2\sin\phi\cos\phi = \sin(2\phi)\). Hence, the integral becomes:\[2 \int_{0}^{\pi/4} \sin(2\phi) \, d \phi.\]This results in:\[-\cos(2\phi) \Bigg|_{0}^{\pi/4} = -\cos(\pi/2) + \cos(0) = 0 + 1 = 1.\]

Step 5: Integrate with Respect to \(\theta\)

Finally, consider the outermost integral:\[\int_{0}^{2\pi} 1 \, d \theta = [\theta]_{0}^{2\pi} = 2\pi.\]

Step 6: Combine the Results

Multiply the results from each step:Inner integral = 4,Middle integral = 1,Outer integral = \(2\pi\).Thus, the final result is:\[4 \times 1 \times 2\pi = 8\pi.\]

Key Concepts

Understanding Triple Integrals

Triple integrals allow you to integrate a function over a three-dimensional space. In spherical coordinates, these are particularly useful for problems involving spheres or spherical regions, like the spherical cap in our problem. The integral you evaluate involves three variables that describe the space:

• \( \rho \), which is the radial distance from the origin.
• \( \phi \), the polar angle measured from the positive z-axis.
• \( \theta \), the azimuthal angle measured in the xy-plane from the positive x-axis.

In our exercise, the triple integral \[ \int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{2}(\rho \cos \phi) \rho^{2} \sin \phi \, d \rho \, d \phi \, d \theta \] is evaluated over a spherical cap, which is a subsection of a sphere determined by these angles and radii. By slicing through this region step-by-step with each integral, we accumulate the function's values throughout that 3D space. Each step simplifies as we integrate through one variable, reducing our problem's dimensionality until we reach a final value.

Exploring the Spherical Cap

A spherical cap is a portion of a sphere cut off by a plane. Picture an ice cream scoop-shaped region on a ball; this is similar to the part of the sphere we are integrating over. The exercise integrates over a spherical cap defined by:

• The radial distance \( \rho \) ranging from 0 to 2, meaning we consider spheres of radius up to 2.
• The angle \( \phi \) ranging from 0 to \( \pi/4 \), carving out the cap's height.
• The full azimuthal sweep \( \theta \) from 0 to \( 2\pi \), which effectively rotates the cap around the z-axis, forming the complete surface.

This configuration effectively sets boundaries for our integration, ensuring we only focus on that specific spherical section. In science and engineering, figuring out the properties of spherical caps helps in calculating volumes and surface areas in fields like meteorology and materials science.

Applying Integration Techniques

Effective integration is vital for solving triple integrals in spherical coordinates. Here, integration techniques make the problem solvable:

• Power Rule: The initial integration with respect to \( \rho \) utilizes the power rule, turning a complicated term \( \rho^3 \cos \phi \sin \phi \) into manageable units by result of \( \frac{\rho^4}{4} \Bigg|_{0}^{2} \).
• Trigonometric Identity: For \( \phi \), identity \( 2\sin\phi\cos\phi = \sin(2\phi) \) simplifies our integral into a more familiar form: \( \sin(2\phi) \), often easier to integrate, resulting in terms of cosine.
• Direct Evaluation: The last stage, involving \( \theta \), is straightforward: the constant factor over the azimuthal angle simplifies to a simple scale multiplier, \( 2\pi \), inscribing the circular path.

Combining these successive integrations explores each region's contribution to the whole, yielding the specific numerical result \( 8\pi \). Understanding and applying each technique helps manage complex integrals, leading to profound insights across various applications in physics and engineering.