Problem 43 - Thomas Calculus

Question

Centroid Find the centroid of the portion of the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) that lies in the first octant.

Step-by-Step Solution

Answer

The centroid is \(\left( \frac{3a}{8}, \frac{3a}{8}, \frac{3a}{8} \right)\).

Step 1: Understand the Problem

We need to find the centroid of a portion of the sphere defined by the equation \(x^2 + y^2 + z^2 = a^2\) that lies in the first octant, meaning all coordinates \((x, y, z)\) are positive.

Step 2: Establish Integral Limits

In the first octant, all coordinates are positive. The limits for \(x\), \(y\), and \(z\) will range from 0 to \(a\) concerning the equation of the sphere \(x^2 + y^2 + z^2 = a^2\). This will naturally imply limits: \(0 \leq x \leq a\), \(0 \leq y \leq \sqrt{a^2 - x^2}\), and \(0 \leq z \leq \sqrt{a^2 - x^2 - y^2}\).

Step 3: Determine the Volume

The volume \(V\) of this region is calculated by integrating the unit volume element \(dV = dx \, dy \, dz\) over the defined limits. So, the integral \( V = \int_0^a \int_0^{\sqrt{a^2 - x^2}} \int_0^{\sqrt{a^2 - x^2 - y^2}} dz \; dy \; dx \) gives us \(\frac{\pi a^3}{6}\).

Step 4: Calculate the Centroid Coordinates

The centroid \((\bar{x}, \bar{y}, \bar{z})\) can be found with the coordinates' formulas for a solid in three dimensions: \(\bar{x} = \frac{1}{V} \int_0^a \int_0^{\sqrt{a^2 - x^2}} \int_0^{\sqrt{a^2 - x^2 - y^2}} x \, dz \, dy \, dx\), \(\bar{y} = \bar{x}\), and \(\bar{z} = \bar{x}\) due to symmetry over all axes. Calculating these integrals gives \(\bar{x} = \bar{y} = \bar{z} = \frac{3a}{8}\).

Step 5: Conclude with the Centroid Coordinates

Thus, the centroid of the sphere portion in the first octant is \(\left( \frac{3a}{8}, \frac{3a}{8}, \frac{3a}{8} \right)\).

Key Concepts

Volume of a Sphere

To understand the concept of the centroid in a portion of a sphere, it's helpful to remember how to find the volume of a sphere itself. A sphere is a three-dimensional figure where every point on its surface is the same distance from the center. For a sphere with radius \(a\), the formula for its volume is given by \(\frac{4}{3} \pi a^3\).

This expression comes from integrating the volume element in spherical coordinates over the entire sphere. However, in this exercise, we are only concerned with a portion of a sphere within the first octant. This region is one-eighth the volume of the whole sphere due to its symmetrical division into eight equal parts.
Finding the volume of this region helps in calculating other properties like the centroid.

Spherical Coordinates

Spherical coordinates are used when dealing with objects like spheres because they simplify the equations involved. Instead of using the usual \((x, y, z)\) three-dimensional coordinate system, spherical coordinates are defined by \((\rho, \theta, \phi)\).

- \(\rho\) is the distance from the origin to the point in space.
- \(\theta\) is the angle in the xy-plane from the positive x-axis.
- \(\phi\) is the angle from the positive z-axis.

These coordinates make it easier to set integral limits over spherical regions. In our problem, setting these coordinates helps to understand the geometry of the sphere and allows us to integrate more easily.

Triple Integration

Triple integration is a powerful tool in calculus used to compute volumes of three-dimensional regions. It involves integrating a function of three variables repeatedly along the \(x\), \(y\), and \(z\) axes. In this exercise, triple integration calculates the volume and centroid coordinates of the sphere's portion in the first octant.

For the sphere in the first octant, this involves calculating:

• The volume \(V\) under set limits: \(\int_0^a \int_0^{\sqrt{a^2 - x^2}} \int_0^{\sqrt{a^2 - x^2 - y^2}} dz \; dy \; dx\).
• The centroid coordinates \((\bar{x}, \bar{y}, \bar{z})\) through similar integrations where the integrands include the respective coordinate multiplied by the volume element.

Through symmetry, each coordinate is equal, simplifying calculations and leading to an understanding of the centroid's position.