Question

A solid is described along with its density function. Find the mass of the solid using spherical coordinates. The cone bounded above \(z=\sqrt{x^{2}+y^{2}}\) and below the plane \(z=1\) with density function \(\delta(x, y, z)=z\).

Short answer

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

Step-by-step solution

Understand the Geometry

The solid is a cone with the apex at the origin. It is bounded above by the surface \(z = \sqrt{x^2 + y^2}\) and below by the plane \(z = 1\). In spherical coordinates, this translates to bounds where \(z\) runs from 1 to \(\sqrt{x^2 + y^2}\).

Convert Coordinates and Density Function

In spherical coordinates, \((x, y, z)\) are expressed as \((\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi)\). The density function becomes \(\delta(\rho, \phi, \theta) = \rho \cos \phi\).

Determine Limits in Spherical Coordinates

The cone boundary is given by \(z=\sqrt{x^2 + y^2}\), which in spherical is \(\rho \cos \phi = \rho \sin \phi\) simplifying to \(\phi = \pi/4\). The bounds are \(1 \leq \rho \leq csc(\phi)\), then \(\phi\) from 0 to \(\pi/4\), and \(\theta\) from 0 to \(2\pi\).

Set Up the Mass Integral in Spherical Coordinates

The mass \(M\) of the solid is given by the triple integral: \[M = \int_0^{2\pi} \int_0^{\pi/4} \int_1^{csc(\phi)} (\rho^2 \cos \phi) \sin \phi \, d\rho \, d\phi \, d\theta\]

Simplify and Evaluate the Integral

The integral simplifies to: \[M = \int_0^{2\pi} d\theta \int_0^{\pi/4} \sin \phi \cos \phi \left[\frac{\rho^3}{3} \right]_1^csc(\phi) \, d\phi\]This becomes:\[M = \frac{2\pi}{3}\int_0^{\pi/4} \csc(\phi)^4 \sin \phi \cos \phi \, d\phi\]Evaluating this integral results in:\[M = \frac{2\pi}{3}(1)\]

Final Calculation

Evaluating gives:\[M = \frac{2\pi}{3}\]

Key concepts

Density Function

When determining the mass of a solid, the density function \(\delta(x, y, z)\) plays a crucial role. Essentially, it tells us how much mass is packed into each unit of space inside the solid. For this specific cone, the density function is given as \(\delta(x, y, z)=z\).
This means that the density increases with the height of the solid. The further you go upwards from the base, the denser the material.

• At the bottom (\(z=1\)), the density is the lowest.
• As you move upwards (\(z\) increases), the density increases proportionally.

Understanding the density helps us calculate the mass more accurately as it changes the contribution of different parts of the solid to the total mass.

Triple Integral

A triple integral is a powerful tool in mathematics to calculate properties like mass by integrating over three-dimensional space. In this problem, the triple integral is used to account for every tiny element's mass in the cone using its density.
The integral has three parts, corresponding to the three dimensions in space:\

• \(\rho\), the radial distance, controls the depth into the cone.
• \(\phi\), the angle from the vertical axis, represents the cone's surface spread.
• \(\theta\), the rotation around the vertical axis, covers the cone's full circular rotation.

Each differential element in the space can be considered as an infinitesimally small block of the solid. In our triple integral, represented as \[ M = \int_0^{2\pi} \int_0^{\pi/4} \int_1^{csc(\phi)} (\rho^2 \cos \phi) \sin \phi \, d\rho \, d\phi \, d\theta \], we combine these elements, integrate over all three dimensions, and thus find the total mass of the cone.

Mass of the Solid

Calculating the mass of the solid involves evaluating the triple integral in the spherical coordinates. Each part of this solid contributes to its entire mass, depending on the density at that particular point. The integrated equality reflects this full account.

• \(M\) represents the total mass of the cone.
• The density and bounds in spherical coordinates are incorporated into the integral.
• After following through with the integration, we find the final mass \(M\) is \(\frac{2\pi}{3}\).

This integral captures how each section of space contributes differently to the mass. The calculation shows that the cone, bounded by specific geometric borders, has its mass distributed according to the given density function, which increases as \(z\) goes up.

Spherical Coordinates Conversion

Spherical coordinates are particularly helpful in problems involving circles and spheres. In this exercise, converting to spherical coordinates simplifies the process of integrating over a cone. The key components in spherical coordinates are:

• \(\rho\), the radial distance from the origin.
• \(\theta\), the angular position around the z-axis.
• \(\phi\), the angle from the positive z-axis.

Their conversion formulas from Cartesian coordinates are \((x, y, z)\) to \((\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi)\).
This conversion helps to better describe objects framed by symmetry, like our cone. In this task, it transitions our boundary conditions into terms more suited for evaluating the integral over the cone's volume.