Problem 21 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Limits and Variables

In spherical coordinates, we have the variables \(\rho\), \(\phi\), and \(\theta\), representing the radial distance, polar angle, and azimuthal angle, respectively. The given limits are \(\rho \) from 0 to \(2\sin\phi\), \(\phi\) from 0 to \(\pi\), and \(\theta\) from 0 to \(\pi\).

Step 2: Set Up the Integral

Write out the integral precisely as it's given: \[\int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{2 \sin \phi} \rho^{2} \sin \phi d\rho \cdot d\phi \cdot d\theta\] The \(\sin \phi\) term appears due to the spherical coordinate Jacobian.

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

Integrate the innermost integral with respect to \(\rho\): \[\int_{0}^{2 \sin \phi} \rho^{2} d\rho = \left[ \frac{\rho^3}{3} \right]_{0}^{2 \sin \phi} = \frac{(2 \sin \phi)^3}{3} = \frac{8 \sin^3 \phi}{3}\]

Step 4: Substitute the Result and Integrate with Respect to \(\phi\)

Substitute back into the integral: \[\int_{0}^{\pi} \int_{0}^{\pi} \frac{8 \sin^3 \phi}{3} \cdot \sin \phi \cdot d\phi \cdot d\theta\] Simplify to: \[\int_{0}^{\pi} \int_{0}^{\pi} \frac{8 \sin^4 \phi}{3} d\phi d\theta\] Now, integrate with respect to \(\phi\): \[\int_{0}^{\pi} \sin^4 \phi \, d\phi\]

Step 5: Evaluate \(\phi\) Integral Using Known Result

The integral of \(\sin^4 \phi\) over \(\phi = 0\) to \(\pi\) is a known result: \[\int_{0}^{\pi} \sin^4 \phi \, d\phi = \frac{3\pi}{8}\] Substitute this back: \[\int_{0}^{\pi} \frac{8}{3} \cdot \frac{3\pi}{8} d\theta = \pi\]

Step 6: Final Integration with Respect to \(\theta\)

Perform the final integration with respect to \(\theta\) over the bounds \(0\) to \(\pi\): \[\int_{0}^{\pi} \pi \, d\theta = \pi[\theta]_{0}^{\pi} = \pi^2\]

Step 7: Verify the Calculation

Ensure that each of the integrations has been performed correctly, especially the substitution and agreement of known integral results. The resulting answer is: \(\pi^2\).

Key Concepts

Triple Integrals

Triple integrals are a fundamental concept in calculus used to integrate functions over three-dimensional regions. This approach enables us to find volumes, masses, and other properties of 3D regions by dividing them into tiny, manageable pieces and summing them up. In the context of spherical coordinates, triple integrals are particularly useful for regions like spheres and cones.

Holding onto the problem at hand, the integral is represented as follows:

• The innermost integral over \( \rho \) corresponds to the radius of our spherical region, totaling from the center out to a boundary defined by \( 2 \sin \phi \).
• The middle integral over \( \phi \) sweeps the polar angle, often ranging from 0 (north pole) to \( \pi \) (south pole).
• The outermost integral over \( \theta \) encapsulates a full circle, with limits from 0 to \( \pi \), completing the integration process in spherical coordinates.

Each integration step simplifies a part of the volume represented by the function inside the integral, ultimately helping find a cumulative value over the 3D space.

Spherical Coordinates

Spherical coordinates are a system where points in three-dimensional space are represented by three parameters: \( \rho \), \( \phi \), and \( \theta \). These correspond to

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

Using spherical coordinates provides advantages in problems with inherent spherical symmetry, allowing simplifications of the integrals.

To convert a Cartesian integral domain into spherical form:

• Recognize the natural bounds for \( \rho \), \( \phi \), and \( \theta \).
• Include the Jacobian of spherical coordinates, \( \rho^2 \sin \phi \), to adjust for the non-linear stretching of space.

This method is particularly beneficial when dealing with problems involving spheres or shells, as it transforms complex Cartesian limits into more straightforward spherical contours.

Calculus Techniques

Calculus provides the toolbox for solving these integral problems efficiently. The integration process progresses by simplifying each integral component separately.

### Integration Steps:In this problem, we used the power of simplification to tackle each integral:

• The first integration performed with respect to \( \rho \) yielded the part \( \frac{8 \sin^3 \phi}{3} \) because we systematically calculated the integral \( \int \rho^2 d\rho \).
• Next, breaking down the integration over \( \phi \) involved multiplying by another \( \sin \phi \), simplifying our task with known integral formulas for functions like \( \sin^n \phi \).
• Finally, integrating the resulting expression in terms of \( \theta \) was simplified due to the constants left from previous steps.

Each piece fits seamlessly by using these techniques, ensuring a methodical evaluation and reducing errors through known results at various steps. This showcases how calculus not only provides solutions but also enhances understanding of geometric and physical properties in 3D space.