Problem 18 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

A spherical star has a radius of 90,000 kilometers. The density of matter in the star is given by \(\rho(r)=\frac{\mathrm{K}}{(r+1)^{3 / 2}}\) kilograms per cubic kilometer, where \(r\) is the distance (in kilometers) from the star's center and \(\mathrm{K}\) is a positive constant. Write out (but do not evaluate) an expression for the total mass of the star. Your answer should contain the constant \(\mathrm{K}\).

Step-by-Step Solution

Answer

The expression for the total mass of the star is \( M = 4\pi K \int_0^{90000} \frac{r^2}{(r+1)^{3 / 2}} dr \).

Step 1: Understand the problem and setup the integral

The density \(\rho(r)\) of the star is given by a function of r, which is the distance from the star's center. The mass (M) can be found by integrating the density function over the volume of the star (V). V is the sphere with a radius of 90,000 kilometers in this case, so limits of integration will be 0 to 90,000 for dr. We use spherical coordinates for this integration, so the volume element dv = r^2 sin(theta) dr d\(\phi\) d\(\theta). We integrate over all angles, so limits are 0 to 2\(\pi\) for d\(\phi\) and 0 to \(\pi\) for d\(\theta).

Step 2: Write the triple integral

The triple integral for the mass of the star becomes \( M = \int_0^{2\pi} \int_0^\pi \int_0^{90000} \frac{K r^2 \sin(\theta)}{(r+1)^{3 / 2}} dr \, d\theta \, d\phi \) . Notice that the integration for the angles (\(d\theta\) and \(d\phi\)) gives a factor of \(4\pi\) (this is the surface area of a unit sphere).

Step 3: Simplify the integral

So the expression for the mass of a star simplifies to \( M = 4\pi K \int_0^{90000} \frac{r^2}{(r+1)^{3 / 2}} dr \) . This is the general expression for the mass of the star as a function of r and K. The question asks us to not evaluate this integral, so this is our final answer.

Key Concepts

Spherical Coordinates

Spherical coordinates play a crucial role in mathematical physics and engineering, especially when dealing with three-dimensional problems involving spherical symmetry. Unlike Cartesian coordinates that specify points using the orthogonal axes x, y, and z, spherical coordinates define a point in space with a distance from a fixed origin and two angles. These three values are commonly notated as \( (r, \theta, \phi) \) where \( r \) is the radius or distance from the origin, \( \theta \) (theta) is the polar angle measured from the positive z-axis, and \( \phi \) (phi) is the azimuthal angle in the xy-plane from the x-axis.

The reason spherical coordinates are so effective in solving problems that involve spheres or spherical objects is because they align with the symmetry of these shapes. Thanks to this alignment, when we set up integrals to calculate properties over a spherical volume — like the mass of a star, as in the exercise — using spherical coordinates simplifies the process. In the case of our star with spherical symmetry, integrating in spherical coordinates enables us to account for every tiny volume element within the star by simply adjusting our radius \( r \) and sweeping through all possible angles \( \theta \) and \( \phi \) from their respective domain limitations.

Tips: Visualizing Spherical Coordinates

For a better conceptual understanding, visualize spherical coordinates as follows:

• Imagine a globe where \( r \) is equivalent to the distance from the Earth's core to a point on the surface.
• The angle \( \theta \) is like the latitude, but it starts from the North Pole instead of the equator.
• The angle \( \phi \) is similar to the longitude but measured around in a full circle from the prime meridian.

This mental model can significantly aid in understanding the setup of integrals in spherical coordinates for real-world applications.

Triple Integral

When it comes to calculating the volume of a three-dimensional object or a property such as mass that is distributed throughout a volume, one often needs to resort to a triple integral. This mathematical tool allows us to evaluate the sum of all tiny contributions throughout a three-dimensional region.

In our exercise, we are concerned with a star's mass, which is a property spread throughout its spherical volume. Therefore, a triple integral is the perfect candidate to sum up all the mass contributions from each infinitesimal volume piece within the star. The triple integral, which we can denote as \( \iiint_V \rho(r) dV \), takes a density function \( \rho(r) \) and integrates it over the entire volume \( V \) of the star.

Understanding the Volume Element in Spherical Coordinates

The volume element \( dV \) in spherical coordinates is particularly convenient for a sphere because it accounts for the change in volume as one moves through the angles and radius. Mathematically, it takes the form \( r^2 \sin(\theta) dr d\phi d\theta \). Here's what each part represents:

• \(\ r^2 \) scales with the square of the radius, accommodating the expanding spherical shells as one moves away from the center.
• \(\ \sin(\theta) \) adjusts for the different 'heights' of latitude as one moves from the poles to the equator of the spherical surface.
• The products of \(\ dr \), \(\ d\phi \), and \(\ d\theta \) ensure that we integrate over all incremental changes in the radius and angles.

By setting proper bounds for \( r \) (from 0 to 90,000 km), \( \theta \) (from 0 to \( \pi \)), and \( \phi \) (from 0 to \( 2\pi \)), we can ensure that the triple integral encompasses the entire volume of the spherical star.

Density Function Integration

In physics, density is defined as mass per unit volume, and this principle is directly applied when we talk about the density function integration. The density function \( \rho(r) \) reflects how the mass is distributed with respect to the distance \( r \) from the center of the star.

When we integrate this density function over the volume of a star, we essentially add up all the tiny masses represented by \( \rho(r) \cdot dV \) to find the total mass. This approach of integration is what allows us to consider variations in density at different radii within the star. As given in the exercise, the function \( \rho(r)=\frac{K}{(r+1)^{3 / 2}} \) signifies that the density decreases with distance from the star's center—this reflects a realistic physical scenario where the core of a star is denser than its outer layers.

Role of the Constant K

The constant \( K \) within our density function is a scaling factor that adjusts the overall magnitude of the density throughout the star. In physical terms, it represents the proportionality between the density and the inverse to the power of \( \frac{3}{2} \) of the radius's increment from the star's center.

Integration of this density function, as demonstrated in the exercise's triple integral, yields an expression for the star's total mass in terms of \( K \). While the exercise specifies not to compute the exact value of the integral and \( K \), understanding how to set up the integral and comprehend the relation of mass and density via integration is foundational in astrophysics and many related fields. Remember that despite not calculating the integral, the formation of the integral itself requires a clear understanding of how mass is distributed within a spherical object and how to mathematically express and process that distribution.