Question

A solid is described along with its density function. Find the mass of the solid using spherical coordinates. The upper half of the unit ball, bounded between \(z=0\) and \(z=\sqrt{1-x^{2}-y^{2}},\) with density function \(\delta(x, y, z)=1\).

Short answer

The mass of the solid is \(\frac{2\pi}{3}\).

Step-by-step solution

Identifying the Region and Density Function

The given solid is the upper half of the unit ball, bounded between \(z=0\) and \(z=\sqrt{1-x^{2}-y^{2}}\). The density function \(\delta(x, y, z)\) is given as 1, indicating constant density. We will integrate over this region using spherical coordinates.

Setting Bounds in Spherical Coordinates

In spherical coordinates, the radius \(\rho\) varies from 0 to 1 (since it is a unit ball), the polar angle \(\theta\) varies from 0 to \(2\pi\) (full rotation around the z-axis), and the azimuthal angle \(\phi\) varies from 0 to \(\frac{\pi}{2}\) (upper half of the sphere, since \(z = \rho \cos \phi\)).

Setup the Mass Integral

The mass \(M\) of the solid can be calculated using the integral \(M = \int \int \int_{V} \delta(x, y, z) \, dV\). In spherical coordinates, this becomes \(M = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} 1 \times \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\).

Integrate with Respect to \(\rho\)

First, we evaluate the integral with respect to \(\rho\):\[ \int_{0}^{1} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_{0}^{1} = \frac{1}{3}. \]

Integrate with Respect to \(\phi\)

Next, we evaluate the integral with respect to \(\phi\):\[ \int_{0}^{\frac{\pi}{2}} \sin \phi \, d\phi = \left[-\cos \phi \right]_{0}^{\frac{\pi}{2}} = 1. \]

Integrate with Respect to \(\theta\)

Finally, we evaluate the integral with respect to \(\theta\):\[ \int_{0}^{2\pi} 1 \, d\theta = \left[ \theta \right]_{0}^{2\pi} = 2\pi. \]

Calculate the Mass

Multiply the results of each integral to find the mass: \[ M = \frac{1}{3} \times 1 \times 2\pi = \frac{2\pi}{3}. \]

Key concepts

Mass of a Solid

Understanding the mass of a solid is crucial in various fields such as physics and engineering. The mass refers to the amount of matter contained in the solid. To calculate it, especially in cases where the shape is complex, one often needs to rely on advanced calculus techniques. These techniques involve integration, which allows us to sum up the contributions of infinitesimally small elements that make up the solid.

In this particular problem, we are dealing with the solid that forms the upper half of a unit ball. To find its mass, we integrate the density function over this defined region. The integration process in spherical coordinates simplifies the computation, making it well-suited for spherical shapes.

Using calculus, we can approach this task effectively by deriving the mass through strategic integration setup, considering both the geometry of the solid and any functions describing its properties.

Unit Ball

The term 'unit ball' refers to a three-dimensional sphere with a radius of one unit. In mathematical problems involving spheres, it frequently acts as a standard test object due to its symmetry and simplicity.

In this problem, we consider only the upper half of the unit ball. This region is bounded by the plane at the equator in the coordinate system, specifically where the z-coordinate is zero. Because of its spherical shape, spherical coordinates are particularly advantageous for dealing with unit balls. This helps us easily translate between the geometric properties of the ball and the required integrals that determine its mass.

The unit ball serves as a familiar example, making it useful for applying and understanding calculus in three-dimensional contexts.

Density Function

A density function in a mathematical context specifies how mass is distributed throughout a given volume. For the problem at hand, the density function is notably simple: \(\delta(x, y, z) = 1\). This means the mass is uniformly distributed throughout the solid.

The uniformity implied by this constant density simplifies calculations significantly. It means that every point within the solid contributes equally to the total mass. Such scenarios make understanding mass calculation more approachable since we don't need to account for variations in density throughout the solid.

Density functions can vary widely in complexity, adding layers to calculations. Here, however, the simplicity allows us to focus purely on mastering integration techniques over a given region.

Triple Integration

Triple integration is an extension of basic integration into multiple dimensions. In a three-dimensional space, it allows us to calculate quantities such as volume, mass, and center of mass, integrating over the x, y, and z dimensions concurrently.

In this exercise, we use triple integration with spherical coordinates, where traditional Cartesian coordinates (x, y, z) are transformed to spherical ones (\(\rho, \phi, \theta\)). This transformation makes integration easier in spherical shapes. The order of integration usually matters. Here, integrating with respect to \(\rho\), \(\phi\), and \(\theta\) sequentially caters to the structure of the upper half sphere from the unit ball.

The final integral formula setup becomes: \[M = \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{1} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\]. This expresses the mass effectively by methodically considering the radial distance, the angle of elevation, and the rotation around the z-axis, simplifying complex three-dimensional integration tasks.