Problem 319 - Calculus Volume 3

Question

A lamina has the shape of a portion of sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) that lies within cone \(z=\sqrt{x^{2}+y^{2}}\) Let \(S\) be the spherical shell centered at the origin with radius \(a\), and let \(C\) be the right circular cone with a vertex at the origin and an axis of symmetry that coincides with the \(z\) -axis. Determine the mass of the lamina if \(\delta(x, y, z)=x^{2} y^{2} z\).

Step-by-Step Solution

Answer

Mass of the lamina is \( \frac{a^7 \pi (4 - \sqrt{2})}{48} \).

Step 1: Understand the Problem

We are given a spherical shell and a cone, and we need to find the mass of the portion of the shell that lies within the cone. The density function given is \( \delta(x, y, z) = x^2 y^2 z \).

Step 2: Parametrize the Surface

We use spherical coordinates to parameterize the surface. Let \( x = a \sin \theta \cos \phi \), \( y = a \sin \theta \sin \phi \), and \( z = a \cos \theta \), where \( 0 \leq \phi < 2\pi \) and \( 0 \leq \theta \leq \frac{\pi}{4} \) since \( z = \sqrt{x^2 + y^2} \) simplifies to \( \theta = \frac{\pi}{4} \).

Step 3: Compute Surface Element

The differential surface area element on a sphere in spherical coordinates is \( dS = a^2 \sin \theta \, d\theta \, d\phi \).

Step 4: Calculate the Function over Surface

Substitute the parameterization into \( \delta(x, y, z) = x^2 y^2 z = (a^2 \sin^2 \theta \cos^2 \phi) (a^2 \sin^2 \theta \sin^2 \phi) (a \cos \theta) \).

Step 5: Integrate to Find the Mass

Integrate the density function over the surface to find the mass: \[M = \int_{\theta=0}^{\frac{\pi}{4}} \int_{\phi=0}^{2\pi} \left( a^5 \sin^4 \theta \cos^2 \phi \sin^2 \phi \cos \theta \right) a^2 \sin \theta \, d\phi \, d\theta.\]This simplifies to:\[M = a^7 \int_{\theta=0}^{\frac{\pi}{4}} \sin^5 \theta \cos \theta \, d\theta \int_{\phi=0}^{2\pi} \cos^2 \phi \sin^2 \phi \, d\phi.\]

Step 6: Integrate over \( \phi \)

Evaluate the integral over \( \phi \):\(\int_{0}^{2\pi} \cos^2 \phi \sin^2 \phi \, d\phi = \frac{1}{8} \int_{0}^{2\pi} \sin^2 (2\phi) \, d\phi = \frac{\pi}{4}.\)

Step 7: Integrate over \( \theta \)

Evaluate the integral over \( \theta \):\(\int_{0}^{\frac{\pi}{4}} \sin^5 \theta \cos \theta \, d\theta = \frac{1}{6} (1 - (\sqrt{2}/2)^6) = \frac{4 - \sqrt{2}}{12}.\)

Step 8: Calculate Total Mass

Combine the results:\[ M = a^7 \left( \frac{4 - \sqrt{2}}{12} \right) \left( \frac{\pi}{4} \right) = \frac{a^7 \pi (4 - \sqrt{2})}{48}. \]

Key Concepts

density function

In mathematics and physics, the concept of a density function is crucial when evaluating quantities like mass or charge distributed over a volume, surface, or line. In this context, the density function is defined as \(\delta(x, y, z) = x^2 y^2 z\), which varies with respect to the coordinates \(x\), \(y\), and \(z\).
This function is spatially dependent, meaning that the density is different at different points within the space described.

• The function describes how mass is distributed over the lamina, which is the portion of the spherical shell in question.
• Density functions are useful because they provide a way to evaluate the distribution of a physical property over a geometrical entity like a surface or volume.
• By integrating the density function over a region, such as the surface of the shell, we can find the total mass contained within that region.

In this problem, the density function influences how we set up our integrals for computing mass.

surface integral

Surface integrals are a natural extension of line integrals to two-dimensional surfaces. They are used to calculate various properties over a surface, such as mass or flux, depending on a given density function. Here, the surface integral is employed to find the mass of the lamina within the spherical shell.

• The purpose of the surface integral is to sum up the infinitesimal contributions of mass distributed over the curved surface of the spherical shell.
• The integration takes into account the surface element, \(dS\), which is determined by parameterizing the surface using spherical coordinates.
• Doing so simplifies the computation because it suits the geometry of the sphere more naturally than Cartesian coordinates would.

Integrating the density function across this surface gives the total mass of the shell portion, factoring in the density at every point.

spherical shell

A spherical shell is a three-dimensional geometrical surface that forms the boundary of a sphere. When dealing with spherical coordinates, understanding the shell layout becomes essential.

• The shell is characterized by all points that lie at a constant distance (radius \(a\)) from a common center, typically the origin in spherical coordinates.
• In this exercise, the portion of the spherical shell lies within a cone defined mathematically by the equation \(z=\sqrt{x^2 + y^2}\).
• Employing spherical coordinates, each point on the shell is represented as \(x = a \sin \theta \cos \phi\), \(y = a \sin \theta \sin \phi\), \(z = a \cos \theta\).

This geometrical setup allows us to analyze surfaces that are symmetrical about a point, which in the case of spheres includes simplifying integration on such surfaces.

mass of lamina

The mass of a lamina refers to the total amount of material or mass contained within a thin layer or surface. To compute this, the challenge involves utilizing both the density function and the surface integral.

• Calculating the mass requires integrating the specified density function \(\delta(x, y, z)\) over the parameterized region of the spherical shell that makes up the lamina.
• The process involves setting up a double integral over the region using the spherical coordinates derived from the shell's geometry, in this case, within the bounds \(0 \leq \phi < 2\pi\) and \(0 \leq \theta \leq \frac{\pi}{4}\).
• After substituting the values and computing the integral, we find that the total mass is given by the complete calculation \(M = \frac{a^7 \pi (4 - \sqrt{2})}{48}\).

Hence, the mass is determined by factoring in both the spatial density distribution and the specific constraints of the region on the spherical surface.