Problem 301 - Calculus Volume 3

Question

Find the mass of a lamina of density \(\xi(x, y, z)=z\) in the shape of hemisphere \(z=\left(a^{2}-x^{2}-y^{2}\right)^{1 / 2}\).

Step-by-Step Solution

Answer

The mass of the lamina is \( \frac{a^4 \pi}{4} \).

Step 1: Understand the Problem

To find the mass of the lamina, we need to integrate the density function over the volume of the hemisphere. The density is given by \( \xi(x, y, z) = z \) and the boundary of the hemisphere is given by \( z = \sqrt{a^2 - x^2 - y^2} \).

Step 2: Set Up the Triple Integral

We set up the triple integral to compute the mass. Since the hemisphere is symmetric and we are using spherical coordinates, it is convenient to integrate in spherical coordinates. We have the limits for \( z \) from \( 0 \) to \( \sqrt{a^2 - x^2 - y^2} \), and for \( x \) and \( y \) we use polar coordinates which gives \( x^2 + y^2 \leq a^2 \), or \( r \) from \( 0 \) to \( a \), and \( \theta \) from \( 0 \) to \( 2\pi \).

Step 3: Convert to Spherical Coordinates

In spherical coordinates, the transformation is \( x = r \cos\theta \), \( y = r \sin\theta \), and \( z = z \). In these coordinates, the differential volume element becomes \( r \, dz \, dr \, d\theta \). The limits for \( r \) are from \( 0 \) to \( a \), \( \theta \) from \( 0 \) to \( 2\pi \), and \( z \) from \( 0 \) to \( \sqrt{a^2 - r^2} \).

Step 4: Set Up the Integral in Converted Coordinates

The mass \( M \) of the lamina can be expressed by the integral: \[ M = \int_0^{2\pi} \int_0^a \int_0^{\sqrt{a^2 - r^2}} z \cdot r \, dz \, dr \, d\theta \]This integral considers the density \( \xi(x, y, z) = z \) and the spherical volume element \( r \, dz \, dr \, d\theta \).

Step 5: Integrate with Respect to \( z \)

First, integrate with respect to \( z \):\[ \int_0^{\sqrt{a^2 - r^2}} z \, dz = \left[ \frac{z^2}{2} \right]_0^{\sqrt{a^2 - r^2}} = \frac{(a^2 - r^2)}{2}\]

Step 6: Integrate with Respect to \( r \)

Now integrate the result with respect to \( r \):\[ \int_0^a \frac{(a^2 - r^2)}{2} \, r \, dr = \frac{1}{2} \int_0^a (a^2 r - r^3) \, dr = \frac{1}{2} \left[ \frac{a^2 r^2}{2} - \frac{r^4}{4} \right]_0^a \]Substitute the limits:\[ = \frac{1}{2} \left( \frac{a^4}{2} - \frac{a^4}{4} \right) = \frac{1}{2} \left( \frac{a^4}{4} \right) = \frac{a^4}{8} \]

Step 7: Integrate with Respect to \( \theta \)

Finally, integrate with respect to \( \theta \):\[ \int_0^{2\pi} \frac{a^4}{8} \, d\theta = \frac{a^4}{8} \cdot \theta \bigg|_0^{2\pi} = \frac{a^4}{8} \cdot 2\pi = \frac{a^4 \pi}{4} \]

Step 8: Write the Final Answer

The mass of the lamina is therefore \( \frac{a^4 \pi}{4} \). This is the result of the computed integral over the volume of the hemisphere.

Key Concepts

Hemisphere

A hemisphere is essentially half of a sphere, like a dome-shaped structure. Imagine slicing a basketball in half, and you'll get the idea. In mathematical terms, a hemisphere is the set of all points
that lie on or beneath one side of a great circle of a sphere. For the problem at hand, the hemisphere is defined as the set of points where the height, or more specifically the z-coordinate, satisfies the boundary equation:
\[ z = \sqrt{a^2 - x^2 - y^2} \].
This equation defines the upper surface of the hemisphere.

• "a" represents the radius of the hemisphere.
• "x" and "y" are coordinates in the plane.

Remember, the hemisphere in this exercise is specifically one where the base lies flat on the xy-plane and the curved part faces upward. This shape's symmetry around the z-axis makes spherical coordinates a logical choice for integration.

Density Function

The density function in a physical context describes how mass is distributed over a given object. For the lamina, the density function is given by
\( \xi(x, y, z) = z \).
This means that the density of the lamina increases as you move higher along the z-axis. The physical intuition here is that regions of the lamina with higher z-coordinates are denser.When performing integration, the density function is crucial. The mass of a small piece of the lamina is calculated by multiplying this density by the volume of the piece.

• The given density function directly ties into how the triple integral is set up.
• Since \( \xi(x, y, z) \) is \( z \), we need to integrate \( z \) over the volume of the hemisphere.

Understanding this function helps grasp why different parts of the hemisphere contribute differently to the overall mass, as denser areas will add more mass.

Spherical Coordinates

Spherical coordinates are designed for dealing with problems involving spheres. They're similar to polar coordinates used in two dimensions but extended to three dimensions. This coordinate system is defined by three values:

• \( r \): the radial distance from the origin to the point.
• \( \theta \): the angle in the xy-plane from the positive x-axis.
• \( \phi \): the angle from the positive z-axis.

In this problem, the use of spherical coordinates is modified to simplify integration over the hemisphere:

• \( r \) bounds from 0 to \( a \), the hemisphere's radius.
• \( \theta \) bounds from 0 to \( 2\pi \), full rotation in the xy-plane.
• \( z \) is bounded from 0 to \( \sqrt{a^2 - r^2} \).

The spherical coordinates make it straightforward to handle integrations over radial distances and rotational symmetry of spheres, reducing much of the complexity found in Cartesian coordinate setups.

Triple Integral

A triple integral is a way to integrate over a three-dimensional region. It allows us to calculate the total sum of a function over that volume.To find the mass of a lamina, we're essentially summing up tiny, infinitesimal masses \( z \cdot r \, dz \, dr \, d\theta \) over the entire space the hemisphere occupies. This specific triple integral setup stems from:

• Inner integral (\( \int_0^{\sqrt{a^2 - r^2}} z \, dz \)): summing up layers from bottom to top of the hemisphere.
• Middle integral (\( \int_0^a r \, dr \)): encapsulates the radial distance.
• Outer integral (\( \int_0^{2\pi} \, d\theta \)): covers the complete angle around the z-axis.

Each step in evaluating the triple integral builds on the work done by the previous, culminating in the calculation of the total mass \( \frac{a^4 \pi}{4} \) as found in the final solution step.