Problem 39 - University Calculus: Early Transcendentals

Question

Set up triple integrals for the volume of the sphere \(\rho=2\) in (a) spherical, (b) cylindrical, and (c) rectangular coordinates.

Step-by-Step Solution

Answer

Set up the triple integrals for each coordinate system as shown in the steps.

Step 1: Understanding the Sphere in Spherical Coordinates

In spherical coordinates, a sphere is defined by the equation \(\rho = 2\), which means the radius from the origin is constant and equal to 2. The spherical coordinate variables are \((\rho, \theta, \phi)\), where \(\theta\) goes from 0 to \(2\pi\) and \(\phi\) goes from 0 to \(\pi\). To find the volume, the integral is set as: \[ \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{2} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \].

Step 2: Setting Up the Integral in Cylindrical Coordinates

The sphere \(\rho=2\) can be expressed in cylindrical coordinates \((r, \theta, z)\) using the equation \(z^2 + r^2 = 4\). To cover the whole sphere, \(r\) ranges from 0 to 2, \(\theta\) from 0 to \(2\pi\), and \(z\) ranges from \(-\sqrt{4 - r^2}\) to \(\sqrt{4 - r^2}\). Thus, the integral becomes: \[ \int_{0}^{2\pi} \int_{0}^{2} \int_{-\sqrt{4 - r^2}}^{\sqrt{4 - r^2}} r \, dz \, dr \, d\theta \].

Step 3: Describing the Sphere in Rectangular Coordinates

In rectangular coordinates, the equation of the sphere \(\rho=2\) becomes \(x^2 + y^2 + z^2 = 4\). The limits for \(z\) are from \(-\sqrt{4 - x^2 - y^2}\) to \(\sqrt{4 - x^2 - y^2}\), while both \(x\) and \(y\) range from \(-2\) to 2, keeping the sphere's volume bounds in mind. Therefore, the integral is: \[ \int_{-2}^{2} \int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \int_{-\sqrt{4 - x^2 - y^2}}^{\sqrt{4 - x^2 - y^2}} \, dz \, dy \, dx \].

Key Concepts

Spherical Coordinates

Spherical coordinates are a 3D coordinate system where the position of a point is determined by three values: the radius \(\rho\), the angle \(\theta\), and the angle \(\phi\). This system is particularly useful for objects like spheres as it inherently incorporates angles and distance from a central origin. In this setup, \(\rho\) is the distance from the origin to the point, \(\theta\) is the azimuthal angle in the xy-plane from the positive x-axis, and \(\phi\) is the polar (or zenith) angle from the positive z-axis.

• The equation \(\rho = 2\) describes a sphere with a radius of 2 centered at the origin.
• To express the volume of this sphere, we use a triple integral.

The integral takes the form:\[ \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{2} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \]The limits for \(\theta\) and \(\phi\) rotate around the z-axis and sweep from pole to pole, respectively. The function \(\rho^2 \sin\phi\) in the integral is the Jacobian determinant, which accounts for the conversion from spherical to Cartesian coordinates.

Cylindrical Coordinates

Cylindrical coordinates are ideal when dealing with shapes that have symmetry around a central axis, such as cylinders. This system uses three values: the radial distance \(r\), the angular coordinate \(\theta\), and the height \(z\).

• Here, \(r\) is the distance from the z-axis, \(\theta\) is the angle in the xy-plane from the positive x-axis, and \(z\) is the height along the z-axis.

For a sphere with radius 2, the expression in cylindrical coordinates becomes\[ z^2 + r^2 = 4 \]To calculate the volume, the limits for \(r\) are from 0 to 2, \(\theta\) from 0 to \(2\pi\), and \(z\) from \(-\sqrt{4 - r^2}\) to \(\sqrt{4 - r^2}\). The integral becomes:\[ \int_{0}^{2\pi} \int_{0}^{2} \int_{-\sqrt{4 - r^2}}^{\sqrt{4 - r^2}} r \, dz \, dr \, d\theta \]The term \(r\) inside the integral arises from the cylindrical to Cartesian transformation, serving as the Jacobian determinant.

Rectangular Coordinates

Rectangular coordinates, also known as Cartesian coordinates, use three mutually perpendicular axes to define the space: x, y, and z. This method is straightforward when dealing with objects with clearly defined edges and corners, but it can still describe spheres.

• In this system, a sphere centered at the origin with radius 2 is described by \(x^2 + y^2 + z^2 = 4\).

The integration for volume in rectangular coordinates can be set using the equation:\[ \int_{-2}^{2} \int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \int_{-\sqrt{4 - x^2 - y^2}}^{\sqrt{4 - x^2 - y^2}} \, dz \, dy \, dx \]The integration limits for each variable ensure the whole sphere is captured, with \(x\) ranging from -2 to 2, \(y\) between the boundaries defined by \(x\), and \(z\) varying according to \(x\) and \(y\). This structure faithfully reconstructs the 3D volume within these bounds.