Problem 27 - Thomas Calculus

Question

The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders are possible and occasionally easier to evaluate. Evaluate the integrals. \(\int_{0}^{2} \int_{-\pi}^{0} \int_{\pi / 4}^{\pi / 2} \rho^{3} \sin 2 \phi d \phi d \theta d \rho\)

Step-by-Step Solution

Answer

The integral evaluates to \(2\pi\).

Step 1: Understanding the Integral

The given integral is in spherical coordinates, where \(\rho\), \(\theta\), and \(\phi\) are the spherical coordinate variables. The order of integration suggests that we evaluate the inner integral over \(\phi\), followed by \(\theta\), and finally \(\rho\).

Step 2: Evaluating the Inner Integral

We start by evaluating the integral with respect to \(\phi\): \[ \int_{\pi/4}^{\pi/2} \rho^3 \sin 2\phi \, d\phi. \] The antiderivative of \(\sin 2\phi\) is \(-\frac{1}{2} \cos 2\phi\). Thus, the integral evaluates to \[-\frac{1}{2} \left[ \rho^3 \cos 2\phi \right]_{\pi/4}^{\pi/2}. \] Substituting the limits of integration yields \[-\frac{1}{2} \rho^3 \left( \cos \pi - \cos \frac{\pi}{2} \right) = -\frac{1}{2} \rho^3 (-1 - 0) = \frac{1}{2} \rho^3. \]

Step 3: Evaluating the Middle Integral

Next, substitute the result from the inner integral into the integral over \(\theta\): \[ \int_{-\pi}^{0} \frac{1}{2} \rho^3 \, d\theta. \] The integral evaluates to \[ \frac{1}{2} \rho^3 \left[ \theta \right]_{-\pi}^{0} = \frac{1}{2} \rho^3 (0 + \pi) = \frac{\pi}{2} \rho^3. \]

Step 4: Evaluating the Outer Integral

Finally, we evaluate the integral with respect to \(\rho\): \[ \int_{0}^{2} \frac{\pi}{2} \rho^3 \, d\rho. \] This integral evaluates as \[ \frac{\pi}{2} \left[ \frac{\rho^4}{4} \right]_{0}^{2} = \frac{\pi}{2} \left( \frac{16}{4} - 0 \right) = \frac{\pi}{2} \times 4 = 2\pi. \]

Key Concepts

Triple Integrals

Triple integrals are advanced calculus tools used to compute the volume under a surface in three-dimensional spaces. These integrals extend the concept of definite integrals from single and double integrals to three coordinates, typically denoted as

• \( x \), \( y \), and \( z \) in Cartesian coordinates
• \( r \), \( \theta \), and \( z \) in cylindrical coordinates
• \( \rho \), \( \theta \), and \( \phi \) in spherical coordinates

Triple integrals are applicable in various fields such as physics and engineering to compute mass, volume, and other spatial properties.
Integrating over a region in a three-dimensional space requires careful evaluation, often done in sequential stages, using a specified order of integration.

Order of Integration

The order of integration in triple integrals dictates the sequence in which the integration is performed over the three coordinate variables. In spherical coordinates, this can be particularly flexible, but sometimes challenging.
For example, in the integral given \[\int_{0}^{2} \int_{-\pi}^{0} \int_{\pi / 4}^{\pi / 2} \rho^{3} \sin 2 \phi \, d \phi \, d \theta \, d \rho,\]we integrate first with respect to \( \phi \), then \( \theta \), and finally \( \rho \). This typical order is influenced by the limits of integration specified for each variable.
Reasons for choosing a specific order can include simplifying the computational process or aligning with the natural bounds of the problem.

• Integrating over \( \phi \) first helps reduce complex trigonometric terms.
• Next, integrating over \( \theta \) eliminates angular dependencies.
• Finally, \( \rho \) allows for simplification of radial distances.

Calculus

Calculus is the mathematical study of change, divided primarily into two subfields: differential calculus and integral calculus. Integral calculus focuses on accumulation, such as areas under curves or volumes under surfaces, using integration as its main operation.
Spherical coordinates are often used in calculus because they simplify the process of integrating over spherical volumes; by converting complex cartesian calculations into easier forms in spherical systems.

• Differential calculus is used to find rates of change or slopes.
• Integral calculus is useful for finding accumulated values, such as area, volume, or mass.

In the context of triple integrals, like the one described here, integral calculus enables us to compute a volume by breaking it down into infinitesimal elements, summing their contributions through integration.