Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-6, evaluate the integral which is given in cylindrical or spherical coordinates, and describe the region \(R\) of integration. \(\int_{0}^{\pi} \int_{0}^{2 \pi} \int_{0}^{a} \rho^{2} \sin \phi d \rho d \theta d \phi\)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{4\pi a^3}{3}\) and describes a hemisphere.

Step 1: Understand the Integral and Coordinate System

The given integral is in spherical coordinates. In spherical coordinates, \(\rho\) represents the radial distance, \(\theta\) the azimuthal angle in the xy-plane, and \(\phi\) the polar angle from the positive z-axis. The integral to evaluate is \( \int_{0}^{\pi} \int_{0}^{2 \pi} \int_{0}^{a} \rho^{2} \sin \phi \, d \rho \, d \theta \, d \phi \).

Step 2: Describe the Region R of Integration

The region \( R \) described by the limits of integration in spherical coordinates is a spherical sector. The radial distance \( \rho \) ranges from 0 to \( a \), \( \theta \) covers the full circle \( 0 \) to \( 2\pi \), and \( \phi \) spans from 0 to \( \pi \), covering the top hemisphere of a sphere.

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

First, integrate with respect to \( \rho \) from 0 to \( a \):\[\int_{0}^{a} \rho^{2} \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^a = \frac{a^3}{3}.\]

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

Next, integrate the result with respect to \( \theta \) from 0 to \( 2\pi \):\[\int_{0}^{2\pi} \frac{a^3}{3} \, d\theta = \left[ \frac{a^3}{3} \theta \right]_0^{2\pi} = \frac{a^3}{3} \times 2\pi = \frac{2\pi a^3}{3}.\]

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

Finally, integrate with respect to \( \phi \) from 0 to \( \pi \) with the \( \sin \phi \) term:\[\int_{0}^{\pi} \frac{2\pi a^3}{3} \sin \phi \, d\phi = \frac{2\pi a^3}{3} \left[ -\cos \phi \right]_0^{\pi} = \frac{2\pi a^3}{3} \left( -(-1) - (1) \right) = \frac{2\pi a^3}{3} \times 2 = \frac{4\pi a^3}{3}.\]

Step 6: Final Result

After completing the integration over all variables, the final result of the integral is \( \frac{4\pi a^3}{3} \), which represents the volume of the hemisphere described by the limits in the spherical coordinate system.

Key Concepts

Triple Integral

Triple integrals are a method used to calculate the volume of a region in space by integrating a function of three variables. In the context of spherical coordinates, these variables are often denoted as \( \rho \,\), \( \phi \,\), and \( \theta \,\). Here's a simple breakdown of how a triple integral works:

• The first integration is performed with respect to \( \rho \,\), which signifies integrating along the radial distance from the origin.
• This is followed by integration with respect to \( \theta \,\), covering the rotation around the z-axis in the xy-plane.
• Finally, the integral is evaluated with respect to \( \phi \,\), describing the arc from the positive z-axis.

By evaluating the given triple integral \( \int_0^{\pi} \int_0^{2\pi} \int_0^a \rho^2 \sin \phi \, d\rho \;d\theta \;d\phi \), we obtain the volume of a certain region, taking into account the entire space described within the given bounds.

Spherical Sector

A spherical sector in mathematics is similar to a pie piece of a sphere. It is determined by a range of angles and a radius within the spherical coordinate system.
In our problem, the spherical sector is defined by:

• \( \rho \) ranging from 0 to \( a \), determining the radius of the sector.
• \( \theta \) from 0 to \( 2\pi \), representing a full circle in the xy-plane.
• \( \phi \) from 0 to \( \pi \), which describes the top hemisphere of the sphere.

This specific sector is actually one hemisphere of a sphere with radius \( a \). When integrating over this region using spherical coordinates, we account for both the volume and shape by observing these limits of \( \rho, \theta, \) and \( \phi \).

Volume of Hemisphere

The volume described in this integral problem is that of a hemisphere, a three-dimensional half-sphere.
A key formula is the volume of a hemisphere:For a sphere, the formula is \( \frac{4}{3}\pi r^3 \). A hemisphere is half of this:\( \text{Volume of hemisphere} = \frac{2}{3}\pi r^3 \)In our case, this can be recognized in the final result of the integral, \( \frac{4\pi a^3}{3} \), which represents twice the hemisphere due to the method of setting the bounds in spherical coordinates.
Our integration over the given region confirms this volume calculation, considering a full spherical sector that accounts for the described hemisphere.

Coordinate Systems

Coordinate systems in mathematics, like spherical coordinates, are crucial for evaluating integrals in multi-dimensional spaces. They provide a convenient way to express points in space by using a combination of radial distances and angles. Spherical coordinates help simplify problems involving symmetry with respect to a sphere, which would otherwise be challenging in Cartesian coordinates.
In spherical coordinates:

• \( \rho \) represents the distance from the origin, which extends outward radially.
• \( \theta \) is the azimuthal angle, indicating direction around the z-axis in the xy-plane.
• \( \phi \) is the polar angle from the z-axis, helping describe vertical tilt.

Such a system is incredibly useful for solving problems involving fields like physics and engineering, where spherical symmetry simplifies the calculations. Importantly, the integration bounds directly reflect the region of space we're interested in, making spherical coordinates an efficient choice for this problem.