Problem 28 - University Calculus: Early Transcendentals

Question

The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders give the same value and are occasionally easier to evaluate. Evaluate the integrals. $$\int_{\pi / 6}^{\pi / 3} \int_{\csc \phi}^{2 \csc \phi} \int_{0}^{2 \pi} \rho^{2} \sin \phi d \theta d \rho d \phi$$

Step-by-Step Solution

Answer

\( \frac{28\pi\sqrt{3}}{9} \)

Step 1: Understand the Integral Setup

The problem involves evaluating a triple integral in spherical coordinates. The order of integration is \( d\theta \), \( d\rho \), \( d\phi \). We integrate with respect to \( \theta \) ranging from 0 to \( 2\pi \), \( \rho \) from \( \csc \phi \) to \( 2 \csc \phi \), and \( \phi \) from \( \pi/6 \) to \( \pi/3 \). The integrand is \( \rho^2 \sin \phi \).

Step 2: Evaluate Inner Integral (dθ)

The inner integral involves \( \theta \) and is of the form \( \int_0^{2\pi} \rho^2 \sin \phi \, d\theta \). Since \( \rho^2 \sin \phi \) does not depend on \( \theta \), it can be treated as a constant over the \( \theta \) integration. Thus, the integral becomes \( \rho^2 \sin \phi \int_0^{2\pi} 1 \, d\theta = \rho^2 \sin \phi \cdot 2\pi \).

Step 3: Evaluate Middle Integral (dρ)

Next, substitute the result from the \( d\theta \) integration and integrate with respect to \( \rho \). We solve \( \int_{\csc \phi}^{2 \csc \phi} \rho^2 \sin \phi \, d\rho \). Factoring out \( \sin \phi \), we have \( 2\pi \sin \phi \int_{\csc \phi}^{2\csc \phi} \rho^2 \, d\rho \), which evaluates to \( 2\pi \sin \phi \left[ \frac{\rho^3}{3} \right]_{\csc \phi}^{2\csc \phi} = 2\pi \sin \phi \left( \frac{(2\csc \phi)^3}{3} - \frac{(\csc \phi)^3}{3} \right) \).

Step 4: Simplify the Integral

The simplification of the \( \rho \) integral gives us \( 2\pi \sin \phi \times \frac{7}{3} \csc^3 \phi = \frac{14\pi}{3} \csc^2 \phi \).

Step 5: Evaluate Outer Integral (dφ)

Integrate the expression with respect to \( \phi \) from \( \pi/6 \) to \( \pi/3 \): \( \frac{14\pi}{3} \int_{\pi/6}^{\pi/3} \csc^2 \phi \, d\phi \). The integral \( \int \csc^2 \phi \, d\phi \) resolves to \( -\cot \phi \), thus: \( \frac{14\pi}{3} \left[ -\cot \phi \right]_{\pi/6}^{\pi/3} = \frac{14\pi}{3} ( -\cot(\pi/3) + \cot(\pi/6) ) \).

Step 6: Final Computation

Finally, we compute \( \cot(\pi/6) = \sqrt{3} \) and \( \cot(\pi/3) = \frac{1}{\sqrt{3}} \). Substituting gives \( \frac{14\pi}{3} ( -\frac{1}{\sqrt{3}} + \sqrt{3} ) = \frac{14\pi}{3} \frac{2\sqrt{3}}{3} = \frac{28\pi\sqrt{3}}{9} \).

Key Concepts

Triple Integral

In mathematics, a triple integral is used to evaluate functions of three variables, typically over a three-dimensional region. For spherical coordinates, these integrals often occur because they are useful for describing problems with symmetry, such as those involving spheres or spherical shells. In spherical coordinates, a point in space is defined using its radius \(\rho\), polar angle \(\phi\), and azimuthal angle \(\theta\). For a triple integral in this system, you would integrate over these three variables.

The triple integral setup in spherical coordinates includes range limits for each variable. Here is what each of these variables signifies in the context of triple integrals:

• \( \rho \): Represents the distance from the origin to a point in space.
• \( \phi \): The angle from the positive \( z \)-axis, typically measured in radians.
• \( \theta \): The angle in the \( xy \)-plane from the positive \( x \)-axis.

When evaluating a triple integral in spherical coordinates, it is critical to consider the order of integration as it determines the region you're integrating over.

Order of Integration

The order of integration in a multiple integral, such as a triple integral, refers to the sequence in which the integrations are performed over each of the variables. In our problem, the order is \( d\theta \), \( d\rho \), and finally \( d\phi \).

Choosing the order of integration can greatly influence the simplicity and feasibility of solving the integral. Generally, the chosen order should simplify the computation by allowing easier integration and utilizing the boundaries of the region. The flexibility in choosing the order allows one to pick paths that eliminate complex expressions earlier in the integration process.

Here’s why the suggested order was used:

• \( d\theta \): Integrating with respect to \( \theta \) first, because the integrand \( \rho^2 \sin \phi \) does not depend on \( \theta \), allowing it to be treated as a constant.
• \( d\rho \): Second because \( \rho \) appears directly in the integrand and is influenced by \( \sin \phi \).
• \( d\phi \): Last because it ties the results together over the specified angular range, completing the integration over the spherical sector.

Evaluation of Integrals

Evaluating the integral involves solving each layer step by step, respecting the order of integration, and understanding the integrand and its limits. Here, the primary focus is on strategy and execution. Let’s see how:

• The first integral over \( \theta \) was straightforward because \( \rho^2 \sin \phi \) was independent of \( \theta \), allowing it to be treated as a constant, yielding an easily computable result over the range \([0, 2\pi]\).
• The second integral over \( \rho \) became a power result because \( \rho^2 \) simplifies to \( \frac{\rho^3}{3} \) when integrated, resulting in clean boundaries after evaluation from \( \csc \phi \) to \( 2 \csc \phi \).
• The final integral over \( \phi \) involved integration of \( \csc^2 \phi \), a standard integral which simplifies using trigonometric identities, leading to the use of the cotangent function.

In every step, carefully applying limits and checking for algebraic simplifications ensures that the integration process is as seamless as possible.

Change of Variables

The concept of change of variables is pivotal when evaluating integrals, particularly in spherical coordinates where transformations simplify the integration. Changing variables refers to altering the coordinate system used to describe the problem, like transitioning from Cartesian to spherical coordinates.

In our specific problem,

• \( dx, dy, dz \)-system requires tedious integration with complex bounds. By changing to spherical coordinates, we can leverage the uniform sphere symmetry.
• The conversion formulas help relate Cartesian to spherical coordinates and vice versa: \( x = \rho \sin \phi \cos \theta \), \( y = \rho \sin \phi \sin \theta \), and \( z = \rho \cos \phi \).
• Volume elements change in this transformation: In spherical coordinates, the volume element \( dV \) becomes \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \), harmonizing perfectly with the integral's geometry.

This transformation is one of the most powerful tools in simplifying and solving complex integrals effectively, making spatial problems more digestible.