Problem 39 - Thomas Calculus

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

Triple integrals are set up in spherical, cylindrical, and rectangular coordinates for the sphere with \\( \rho = 2 \\\).

Step 1: Understand the Problem

We need to set up triple integrals to find the volume of a sphere with radius 2. We'll do this in three different coordinate systems: spherical, cylindrical, and rectangular.

Step 2: Spherical Coordinates Setup

In spherical coordinates, the sphere \( \rho = 2 \) is described by \( 0 \leq \phi \leq \pi \), \( 0 \leq \theta \leq 2\pi \), and \( 0 \leq \rho \leq 2 \). The volume element is \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \). Thus, the triple integral is set up as:\[ \int_0^{2\pi} \int_0^{\pi} \int_0^2 \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta. \]

Step 3: Cylindrical Coordinates Setup

In cylindrical coordinates, a sphere of radius 2 is described by \( r^2 + z^2 \leq 4 \). The bounds for \( \theta \) are \( 0 \leq \theta \leq 2\pi \); for \( r \) are \(0 \leq r \leq 2 \); and for \( z \) are \( -\sqrt{4 - r^2} \leq z \leq \sqrt{4 - r^2} \). The volume element is \( dV = r \, dr \, d\theta \, dz \). The setup for the integral is:\[ \int_0^{2\pi} \int_0^2 \int_{-\sqrt{4 - r^2}}^{\sqrt{4 - r^2}} r \, dz \, dr \, d\theta. \]

Step 4: Rectangular Coordinates Setup

In rectangular coordinates, the sphere \( x^2 + y^2 + z^2 \leq 4 \) is bounded by \( -2 \leq x \leq 2 \), \( -\sqrt{4 - x^2} \leq y \leq \sqrt{4 - x^2} \), and \( -\sqrt{4 - x^2 - y^2} \leq z \leq \sqrt{4 - x^2 - y^2} \). The volume element is \( dV = dx \, dy \, dz \). Therefore, the integral is set up as:\[ \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{-\sqrt{4-x^2-y^2}}^{\sqrt{4-x^2-y^2}} dz \, dy \, dx. \]

Step 5: Conclusion

The triple integrals have been set up for each coordinate system. The integrals are not solved but set up to find the volume of the sphere with radius 2.

Key Concepts

Spherical Coordinates

Spherical coordinates are a 3D coordinate system that helps simplify complex problems, especially those with symmetry about a point, like spheres. In this system, we use three parameters to describe any point in space:

• \(\rho\): the distance from the origin to the point.
• \(\phi\): the angle between the positive z-axis and the line connecting the origin to the point.
• \(\theta\): the angle between the positive x-axis and the projection of the line connecting the origin to the point onto the xy-plane.

For the sphere with a radius of 2, the equation \(\rho = 2\) implies it has a constant radius in all directions. The bounds for the spherical coordinates are given as: - \(0 \leq \phi \leq \pi\)- \(0 \leq \theta \leq 2\pi\)- \(0 \leq \rho \leq 2\) The differential volume element in spherical coordinates is \(dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\), which represents an infinitesimally small chunk of the sphere. The triple integral is set up with these limits to compute the volume of the sphere.

Cylindrical Coordinates

Cylindrical coordinates are useful in problems involving cylinders or rotational symmetry about an axis. This system includes three parameters:

• \(r\): radius of the projection of the point onto the xy-plane.
• \(\theta\): the angle from the positive x-axis to the point's projection on the xy-plane.
• \(z\): the height above the xy-plane.

For the sphere of radius 2, the spherical equation \(r^2 + z^2 \leq 4\) describes the slice of the sphere in this system. The integral setup requires working through: - \(0 \leq \theta \leq 2\pi\) for complete circular symmetry.- \(0 \leq r \leq 2\) to cover up to the sphere's radius.- \(-\sqrt{4 - r^2} \leq z \leq \sqrt{4 - r^2}\) to ensure points remain inside the sphere.The volume element here is \(dV = r \, dr \, d\theta \, dz\). This ensures that each piece contributes correctly to the volume calculation.

Rectangular Coordinates

Rectangular coordinates, also known as Cartesian coordinates, are the most straightforward system but often not the simplest for spherical objects. They describe a point in space using:

• \(x\): the distance along the x-axis.
• \(y\): the distance along the y-axis.
• \(z\): the distance along the z-axis.

For a sphere with radius 2, we set the equation \(x^2 + y^2 + z^2 \leq 4\) to express the constraint. The bounds become more complex since they need to account for all regions inside the sphere:- \(-2 \leq x \leq 2\) covers the diameter along the x-axis.- \(-\sqrt{4 - x^2} \leq y \leq \sqrt{4 - x^2}\) varies as \(x\) changes to stay within the circular cross-section.- \(-\sqrt{4 - x^2 - y^2} \leq z \leq \sqrt{4 - x^2 - y^2}\) ensures points stay within the sphere.The differential volume is \(dV = dx \, dy \, dz\), summing up the tiny volume units across the space to encapsulate the sphere's interior accurately.