Problem 42 - APEX Calculus

Question

A region is space is described. Set up the triple integrals that find the volume of this region using rectangular, cylindrical and spherical coordinates, then comment on which of the three appears easiest to evaluate. The cube enclosed by the planes \(x=0, x=1, y=0\) \(y=1, z=0\) and \(z=1\). (Hint: in spherical, use order of integration \(d \rho d \varphi d \theta\).)

Step-by-Step Solution

Answer

The triple integral in rectangular coordinates is easiest to evaluate due to constant limits and alignment with the region's symmetry.

Step 1: Set Up Triple Integral in Rectangular Coordinates

The volume of the cube is calculated by integrating over its entire region. For a cube with boundaries from 0 to 1 along the x, y, and z axes, the triple integral in rectangular coordinates is set up as follows:\[\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} dz \; dy \; dx\]The order of integration is with respect to z, then y, and finally x. This region is straightforward to integrate since all limits are constants.

Step 2: Set Up Triple Integral in Cylindrical Coordinates

In cylindrical coordinates, the transformation from rectangular coordinates is given by \(x = r \cos(\theta)\), \(y = r \sin(\theta)\), and \(z = z\). The volume integral involves an additional \(r\) in the integrand due to the Jacobian of the transformation:\[\int_{0}^{1} \int_{0}^{\pi/4} \int_{0}^{1} r \, dz \, dr \, d\theta\]The limits for \(r\) and \(z\) are 0 to 1, and for \(\theta\) the range is 0 to \(\pi/4\) since the cube lies in the first quadrant where both x and y span from 0 to 1.

Step 3: Set Up Triple Integral in Spherical Coordinates

In spherical coordinates, the transformation is \(x = \rho \sin(\varphi) \cos(\theta)\), \(y = \rho \sin(\varphi) \sin(\theta)\), \(z = \rho \cos(\varphi)\). The volume element includes \(\rho^2 \sin(\varphi)\) as part of the Jacobian:\[\int_{0}^{\frac{\pi}{4}} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{3}} \rho^2 \sin(\varphi) \, d\rho \, d\varphi \, d\theta\]The limits for \(\rho\) are 0 to \(\sqrt{3}\), for \(\varphi\) from 0 to \(\pi/4\), and for \(\theta\) from 0 to \(\pi/4\) considering the bounded region of the cube.

Step 4: Compare Ease of Integration

Evaluating the triple integral in rectangular coordinates is the simplest, as all the limits are constants and easily integrable from fundamental theorems of integration. Cylindrical and spherical coordinates introduce trigonometric functions which complicate the integrals. Additionally, the symmetry of the cube is more naturally aligned with rectangular coordinates.

Key Concepts

rectangular coordinates

When we visualize space, we often think in rectangular coordinates because they align perfectly with our typical 3D perspective. Each point in space is identified using three perpendicular axes: x, y, and z. Rectangular coordinates are straightforward, especially when dealing with cubic shapes like the one in the exercise.

In the context of this exercise, the cube's boundaries lie between 0 and 1 along each axis. To compute its volume using triple integrals, we simply integrate a constant function over these constant limits. This is expressed mathematically as:

• Integrate with respect to z first: \(\int_{0}^{1} dz \\)
• Then integrate with respect to y: \(\int_{0}^{1} dy \\)
• Finally, integrate with respect to x: \(\int_{0}^{1} dx \\)

Breaking down the region in this manner takes advantage of the cube's uniform dimensions. Each integral handles a different dimension, simplifying the evaluation process significantly.

cylindrical coordinates

Cylindrical coordinates are particularly useful for problems involving symmetry around an axis, and they extend polar coordinates from two dimensions to three. In cylindrical coordinates, each point in space is defined by:

• \(r\): the radial distance from the origin in the xy-plane.
• \(\theta\): the angle around the z-axis from the positive x-axis.
• \(z\): the height above the xy-plane.

In our cube example, the conversion from rectangular to cylindrical coordinates introduces complexity due to additional terms in the integral. For instance, the volume element now includes an \(r\) term. The integrals become:

• \(\int_{0}^{1} r \, dz\)
• \(\int_{0}^{1} \, dr\)
• \(\int_{0}^{\pi/4} \, d\theta\)

The \(\theta\) limit accounts for any rotations that cause the cube to lie within 0 to \(\pi/4\), the region's first octant. Although cylindrical coordinates can seem more challenging, they're invaluable when dealing with circular components within bounded regions.

spherical coordinates

Spherical coordinates are ideal for volumes centered around a point, particularly those with spherical surfaces. They describe a point with:

• \(\rho\): the radial distance from the origin.
• \(\varphi\): the angle from the positive z-axis.
• \(\theta\): the angle around the z-axis, similar to in cylindrical coordinates.

For this exercise, transforming the cube description into spherical coordinates adds complexity. Because this form is suited for spherical bodies, fitting a cube requires careful setting of limits. The volume element is modified to include \(\rho^2 \sin(\varphi)\), accounting for changes in volume density as you move from the center outward:

• \(\int_{0}^{\sqrt{3}} \rho^2 \, d\rho\)
• \(\int_{0}^{\pi/4} \sin(\varphi) \, d\varphi\)
• \(\int_{0}^{\pi/4} \, d\theta\)

Converting to spherical coordinates for a cubic region shows less inherent symmetry compared to rectangular coordinates, thus making the integral harder to evaluate. However, understanding spherical coordinates is crucial for tackling problems involving spheres or spherical segments.