Question

Use triple iterated integrals to find the indicated quantities. Center of mass of that part of the solid sphere \(\\{(x, y, z):\) \(\left.x^{2}+y^{2}+z^{2} \leq a^{2}\right\\}\) that lies in the first octant, assuming that it has constant density

Short answer

Center of mass is \((\frac{3a}{4\pi}, \frac{3a}{4\pi}, \frac{3a}{4\pi})\)."

Step-by-step solution

Understand the Geometry and Setup

The problem involves a solid sphere of radius \(a\) that is located in the first octant of three-dimensional space. The first octant refers to the region where \(x, y,\) and \(z\) are all non-negative. We need to find the center of mass of this spherical segment assuming a constant density. This suggests using spherical coordinates for the integration due to the symmetry of the sphere.

Describe the Coordinate System

In spherical coordinates, any point \((x, y, z)\) can be described by \((\rho, \theta, \phi)\), where \(\rho\) is the distance from the origin, \(\theta\) is the azimuthal angle in the xy-plane, and \(\phi\) is the polar angle from the positive z-axis. The transformations are \(x = \rho \sin \phi \cos \theta\), \(y = \rho \sin \phi \sin \theta\), and \(z = \rho \cos \phi\). The limits for \(\rho\) are from 0 to \(a\), \(\theta\) is from 0 to \(\frac{\pi}{2}\) (as it is first octant), and \(\phi\) from 0 to \(\frac{\pi}{2}\).

Set up Integral for Mass

The mass \(M\) of the region can be calculated as \(M = \iiint_\text{first octant} \delta \, dV\). Given constant density \(\delta\), it simplifies to \(M = \delta \iiint_\text{first octant} dV\). Converting to spherical coordinates, this integral becomes \(M = \delta \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^a \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\), where \(dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\).

Evaluating the Integral for Mass

Evaluate the integral for mass: \[ \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^a \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta = \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \left. \frac{\rho^3}{3} \right|_0^a \sin \phi \, d\phi \, d\theta = \frac{a^3}{3} \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \sin \phi \, d\phi \, d\theta \]Calculating the inner integrals:\[ = \frac{a^3}{3} \int_0^{\frac{\pi}{2}} \left. -\cos \phi \right|_0^{\frac{\pi}{2}} \, d\theta \]\[ = \frac{a^3}{3} \int_0^{\frac{\pi}{2}} (1 - 0) \, d\theta = \frac{a^3}{3} \cdot \frac{\pi}{2} = \frac{\pi a^3}{6} \]Thus, the mass of the region is \(M = \delta \frac{\pi a^3}{6}\).

Calculate Coordinates of the Center of Mass

To find the center of mass \((\bar{x}, \bar{y}, \bar{z})\), use the formulas \(\bar{x} = \frac{1}{M} \iiint x \delta \, dV\), \(\bar{y} = \frac{1}{M} \iiint y \delta \, dV\), and \(\bar{z} = \frac{1}{M} \iiint z \delta \, dV\). Given the symmetry and limits, each integral is identical; hence, we only need to calculate \(\bar{x}\).

Calculate \(\bar{x}\)

\[ \bar{x} = \frac{1}{M} \delta \iiint x \, dV = \frac{1}{\frac{\pi a^3}{6}} \delta \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^a (\rho \sin \phi \cos \theta) \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]\[ = \frac{6\delta}{\pi a^3} \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^a \rho^3 \sin^2 \phi \cos \theta \, d\rho \, d\phi \, d\theta \]Evaluate the \(\rho\) integral:\[ = \frac{6\delta}{\pi a^3} \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \left. \frac{\rho^4}{4} \right|_0^a \, \sin^2 \phi \, \cos \theta \, d\phi \, d\theta \]\[ = \frac{6\delta}{\pi a^3} \cdot \frac{a^4}{4} \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \sin^2 \phi \cos \theta \, d\phi \, d\theta \]\[ = \frac{6\delta a}{4\pi} \int_0^{\frac{\pi}{2}} \cos \theta \, d\theta \int_0^{\frac{\pi}{2}} \sin^2 \phi \, d\phi \]Calculating each integral, you find \(\bar{x} = \frac{6a}{8\pi}\) after cancelling out terms.

Symmetrical Property Application for \(\bar{y}\) and \(\bar{z}\)

By symmetry, in the first octant, the center of mass should be equally divided among the directions. Thus, \(\bar{y} = \bar{x}\) and \(\bar{z} = \bar{x}\).

Final Values of Center of Mass

The center of mass coordinates are \((\bar{x}, \bar{y}, \bar{z}) = \left(\frac{6a}{8\pi}, \frac{6a}{8\pi}, \frac{6a}{8\pi}\right)\). Simplify further to \((\frac{3a}{4\pi}, \frac{3a}{4\pi}, \frac{3a}{4\pi})\).

Key concepts

Center of Mass

The center of mass of an object is a crucial concept in physics and math. It represents the point at which the entire mass of a body can be considered to be concentrated. When dealing with symmetrical objects like spheres, calculations become simpler. This is because the mass distribution is uniform.

For this specific exercise, we are dealing with a portion of a sphere that lies within the first octant. The sphere has a constant density which simplifies our calculations. We are calculating the center of mass using triple iterated integrals. This method is particularly effective because it captures the three-dimensional nature of the object. Here, we recognize the common traits of a uniform sphere. Thus, the center of mass will lie on the line bisecting the object. Since we're only considering the first octant, symmetry helps. It implies that each coordinate of the center of mass (\(\bar{x}, \bar{y}, \bar{z}\)) will be identical, facilitating our calculations and giving a clearer understanding of the distribution of mass.

Spherical Coordinates

Spherical coordinates provide a natural way to describe points in 3D space. They are especially useful for problems with spherical symmetry, like this exercise. In this coordinate system, any point \((x, y, z)\) is represented by three parameters: \(\rho\) (the radius), \(\theta\) (the azimuthal angle in the xy-plane), and \(\phi\) (the polar angle from the positive z-axis).

The transformations are straightforward:

• \(x = \rho \sin \phi \cos \theta\)
• \(y = \rho \sin \phi \sin \theta\)
• \(z = \rho \cos \phi\)

For the sphere in the first octant, the limits of integration are adjusted to capture only this region. \(\rho\) ranges from 0 to \(a\) (the radius of the sphere), \(\theta\) covers \((0, \frac{\pi}{2})\), and \(\phi\) also spans \((0, \frac{\pi}{2})\). This restricted range ensures we partake only in calculations for the first octant. Spherical coordinates simplify integrating over volumes like spheres, allowing us to harness symmetry and streamline computations.

Constant Density

Assuming constant density \(\delta\) in the part of the sphere we are considering makes calculations more manageable. Specifically, it means that the density doesn't change throughout the volume. This constancy allows us to pull \(\delta\) outside the integral when calculating mass. As the density does not vary, we treat it as a constant multiplier.

When we calculate the mass of the sphere's segment using spherical coordinates, this constant density underpins the method. It allows for straightforward multiplication with the volume integral. This simplification is vital as it keeps integral calculations neat and focused only on the geometrical properties. \(M = \delta \iiint\_\text{first octant} dV\), highlights how \(\delta\) factors into mass calculations. Ultimately, constant density preserves proportionality in calculations - crucial when identifying where the center of mass lies.

First Octant

The first octant in a three-dimensional space refers to the region where all three coordinates \((x, y, z)\) are non-negative. In simpler terms, it's the section of the coordinate system where we only consider positive values for all axes. Visualizing the octant helps set boundaries for problems like this, ensuring we only tackle the necessary volume of interest.

In this exercise, we're dealing with a part of a sphere that falls in the first octant, bounded by \(x, y, z \geq 0\). This restriction shapes our integration limits within spherical coordinates. Spherical coordinates elegantly fit this problem since they allow us to tailor limits precisely. The first octant's specificity aids in making integrals finite and focused. Since we know the sphere's symmetry and density are constant, this limiting factor further simplifies computations, leading directly to the conclusion that mass and center align within this defined three-dimensional corner.