Problem 26 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{\pi}{6}(2\sqrt{2} - 1)\).

Step 1: Identify the Given Integral

The given integral is specified in spherical coordinates with the limits \( \rho \) from 0 to \( \sec \phi \), \( \phi \) from 0 to \( \pi/4 \), and \( \theta \) from 0 to \( 2\pi \). The integrand is \( \rho \cos\phi \times \rho^{2} \sin \phi \).

Step 2: Simplify the Integrand

The integrand \( \rho \cos \phi \times \rho^2 \sin \phi \) simplifies to \( \rho^3 \cos \phi \sin \phi \). Thus, the complete integral is now:\[\int_{0}^{2\pi} \int_{0}^{\pi/4} \int_{0}^{\sec \phi} \rho^3 \cos \phi \sin \phi \, d\rho \, d\phi \, d\theta\]

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

Integrate \( \rho^3 \) with respect to \( \rho \) from 0 to \( \sec \phi \). This ends by evaluating:\[\int_{0}^{\sec \phi} \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_{0}^{\sec \phi} = \frac{1}{4}(\sec \phi)^4\]

Step 4: Substitute the \( \rho \) Integral Result

Substitute the result from Step 3 back into the integral:\[\int_{0}^{2\pi} \int_{0}^{\pi/4} \frac{1}{4}(\sec \phi)^4 \cos \phi \sin \phi \, d\phi \, d\theta\]

Step 5: Simplify Using Trigonometric Identities

Note that \( (\sec \phi)^4 \cos \phi = \frac{1}{\cos^4 \phi} \cdot \cos \phi = \sec^3 \phi \). Therefore, the integral reduces to:\[\int_{0}^{2\pi} \int_{0}^{\pi/4} \frac{1}{4} \sec^3 \phi \sin \phi \, d\phi \, d\theta\]

Step 6: Integrate with Respect to \( \phi \)

Integrate \( \frac{1}{4} \sec^3 \phi \sin \phi \) with respect to \( \phi \). The antiderivative is evaluated as:\[\int_{0}^{\pi/4} \frac{1}{4} \sec^3 \phi \sin \phi \, d\phi = \frac{1}{4} \left[ \frac{1}{3} \sec^3 \phi \right]_{0}^{\pi/4} = \frac{1}{12} \left( \sqrt{2}^3 - 1 \right)\]This evaluates to \(\frac{1}{12}(2\sqrt{2} - 1)\).

Step 7: Integrate with Respect to \( \theta \)

Finally, since the integration with respect to \( \theta \) is straightforward because there is no \( \theta \) in the integrand, it just multiplies the result by the range \( 2\pi \):\[2\pi \times \frac{1}{12}(2\sqrt{2} - 1)\]

Step 8: Compute the Final Result

Calculate the final result to get:\[\frac{\pi}{6}(2\sqrt{2} - 1)\]

Key Concepts

Triple Integral

When we deal with a triple integral, we are essentially calculating a volume in a three-dimensional space. These integrals extend the idea of double integrals (which calculate area) to three dimensions. In the case of spherical coordinates, the triple integral allows us to accumulate a sum of values over a volume, reflecting their locations in a spherical space.
Triple integrals require boundaries for three variables. In spherical coordinates, these variables are usually denoted as \( \rho \), \( \phi \), and \( \theta \).

• \( \rho \) is the radial distance from the origin.
• \( \phi \) is the angle from the positive z-axis.
• \( \theta \) is the azimuthal angle, the same as in polar coordinates.

In the original exercise, we found the limits for integration based on the context given. The process involves integrating sequentially with respect to each variable, starting with \( \rho \), followed by \( \phi \), and finally \( \theta \). This helps in evaluating the contribution of each infinitesimal volume element within the specified boundaries.

Integration in Spherical Coordinates

Spherical coordinates are a natural fit when the problem at hand has radial symmetry. This system utilizes the radius and angles to locate a point in space, making it useful in physics and engineering, particularly for problems involving spheres or circular symmetry.
In spherical coordinates, the differential volume element, expressed in terms of \( d\rho \), \( d\phi \), and \( d\theta \), is given by \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \). This formula incorporates the spherical geometry into the integration process.

The exercise integrates the function over a range of spherical coordinates. To evaluate this, follow these steps:

• Identify the integration boundaries for \( \rho \), \( \phi \), and \( \theta \) based on the volume of interest.
• Set up the integral using the spherical differential volume element, ensuring to multiply the integrand by \( \rho^2 \sin \phi \).

In our example, the integration step-by-step starts with the radial variable \( \rho \), which simplifies the process once combined with the angular integrations. This approach simplifies calculations involving spherical shaped regions.

Trigonometric Identities

Trigonometric identities play a crucial role in simplifying integrals, especially those involving spherical coordinates. In spherical coordinate integration, trigonometric functions often appear due to the geometry of the sphere.
For example, in our integration process, we encountered the identity involving secant: \( \sec \phi = \frac{1}{\cos \phi} \). Applying these identities is helpful because:

• They reduce complex expressions into simpler forms, making integrations more manageable.
• They aid in transforming given functions so that antiderivatives easier to find.

In our exercise, noticing that \( \sec^3 \phi \cos \phi \) simplifies to \( \sec^3 \phi \) using trigonometric identities, simplifies the integration significantly.
The importance of knowing these identities cannot be overstated, as they let us rewrite difficult integrands into expressions we can integrate using standard techniques. As a result, complex-looking problems become surprisingly straightforward.