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Chapter 7: Problem 5

Find the centroid of the half-ball of radius \(a\) in \(\mathbb{R}^{3}\) that lies inside the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) and above the \(x y\) -plane.

Short Answer

Expert verified
The centroid of the half-ball is \((0, 0, \frac{3}{8} a)\)."

Step by step solution

01

Understanding the Problem

The problem asks us to find the centroid of a half-ball of radius \( a \) in 3D space. This half-ball is defined by a sphere \( x^2 + y^2 + z^2 = a^2 \) that lies above the \( xy \)-plane.
02

Setting Up the Integral

The centroid \((\bar{x}, \bar{y}, \bar{z})\) of a volume can be found using integrals: \(\bar{x} = \frac{1}{V}\int x \, dV\), \(\bar{y} = \frac{1}{V}\int y \, dV\), \(\bar{z} = \frac{1}{V}\int z \, dV\), where \(V\) is the volume of the half-ball. Because of symmetry about the \(z\)-axis, we can immediately conclude that \(\bar{x}=0\) and \(\bar{y}=0\).
03

Calculating the Volume

The volume of the half-ball is half the volume of the full sphere, which is \(\frac{1}{2} \cdot \frac{4}{3}\pi a^3 = \frac{2}{3}\pi a^3\).
04

Expressing dV in Spherical Coordinates

To calculate \(\bar{z}\), we use spherical coordinates: \( x = \rho \sin \theta \cos \phi \), \( y = \rho \sin \theta \sin \phi \), \( z = \rho \cos \theta \). The differential volume element is \( dV = \rho^2 \sin \theta \, d\rho \, d\theta \, d\phi \).
05

Setting Up the Integral for \(\bar{z}\)

The limits for \( \rho \) are \( 0 \) to \( a \), for \( \theta \) are \( 0 \) to \(\frac{\pi}{2}\) (since it's a half-ball above the \(xy\)-plane), and for \( \phi \) are \( 0 \) to \(2\pi\). The integral to find \( \bar{z} \) becomes \( \bar{z} = \frac{1}{V} \int_0^{2\pi} \int_0^{\pi/2} \int_0^a (\rho \cos \theta) \rho^2 \sin \theta \, d\rho \, d\theta \, d\phi \).
06

Solving the Integral

First integrate with respect to \( \rho \):\[\int_0^a \rho^3 \, d\rho = \left[ \frac{\rho^4}{4} \right]_0^a = \frac{a^4}{4}\]Second, integrate with respect to \( \theta \):\[\int_0^{\pi/2} \cos \theta \sin \theta \, d\theta = \left[ \frac{\sin^2 \theta}{2} \right]_0^{\pi/2} = \frac{1}{2}\]Finally, integrate with respect to \( \phi \):\[\int_0^{2\pi} \, d\phi = 2\pi\]Therefore, the result of the triple integral is \( \frac{a^4}{4} \cdot \frac{1}{2} \cdot 2\pi = \frac{\pi a^4}{4} \).
07

Computing \(\bar{z}\)

Using the formula for \(\bar{z}\), we get:\[\bar{z} = \frac{1}{\frac{2}{3} \pi a^3} \cdot \frac{\pi a^4}{4} = \frac{3}{8} a\]
08

Conclusion: Reporting the Centroid

The centroid of the half-ball is \((0, 0, \frac{3}{8} a)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spherical coordinates
To solve problems involving regions like the half-ball above the xy-plane, spherical coordinates are incredibly useful.The spherical coordinate system helps us to describe a point in 3D space using three parameters:
  • \( \rho \): the radial distance from the origin;
  • \( \theta \): the angle between the positive z-axis and the line formed from the origin to the radial projection on the xy-plane;
  • \( \phi \): the angle between the positive x-axis and the projection of the line on the xy-plane.
These parameters are related to Cartesian coordinates by the equations: \[ x = \rho \sin \theta \cos \phi, \quad y = \rho \sin \theta \sin \phi, \quad z = \rho \cos \theta. \]Spherical coordinates decouple the dependencies that exist between x, y, and z in Cartesian coordinates. This is particularly useful when dealing with spherical regions because the bounds of integration can be aligned with the geometry of the problem, making integrals easier to evaluate. For example, in our exercise, the radial coordinate \( \rho \) ranges from 0 to \( a \), aligning neatly with the sphere's radius.
Triple integrals
In calculus, triple integrals allow us to calculate volumes, centroids, and other properties of 3D regions. For the half-ball problem, we calculate the centroid using triple integrals.A triple integral over a region can be expressed as \[ \int \int \int_E f(x, y, z) \, dV, \]where \( dV \) represents the volume element in 3D space. By converting to spherical coordinates, we express \( dV \) as \[ dV = \rho^2 \sin \theta \, d\rho \, d\theta \, d\phi. \]The function \( f(x, y, z) \) changes depending on the property we wish to calculate. For finding the centroid, we use the functions \( f = x \), \( f = y \), and \( f = z \), one at a time, in the integrals to determine \( \bar{x} \), \( \bar{y} \), and \( \bar{z} \).The limits of integration for \( \rho \), \( \theta \), and \( \phi \) are determined by the boundaries of the region鈥攈ere, they correspond to half of the sphere above the xy-plane. These integrals are instrumental in calculating the centroid by integrating over the entire region.
Symmetry in calculus
Symmetry can greatly simplify many calculus problems, including those dealing with centroids and integrals.In the exercise provided, symmetry plays a crucial role in immediately determining that \( \bar{x} \) and \( \bar{y} \) are both zero. This is because the half-ball is symmetric about the z-axis. Hence, for every mass element on one side of the z-axis, there is an equivalent element on the opposite side, perfectly balancing each other out in the x and y directions.This simplifies the problem significantly since we only need to calculate \( \bar{z} \), the component that is not affected by symmetry in this orientation. By recognizing the symmetric nature of the problem, we focus only on the calculations necessary for \( \bar{z} \), using triple integrals in spherical coordinates to achieve this. This not only reduces computation but also aids in understanding how geometry and calculus can harmonize through the concept of symmetry.

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Most popular questions from this chapter

In this exercise, we evaluate the improper one-variable integral \(\int_{-\infty}^{\infty} e^{-x^{2}} d x\) by following the unlikely strategy of relating it to an improper double integral that turns out to be more tractable. Let \(a\) be a positive real number. (a) Let \(R_{a}\) be the rectangle \(R_{a}=[-a, a] \times[-a, a] .\) Show that: $$ \iint_{R_{a}} e^{-x^{2}-y^{2}} d x d y=\left(\int_{-a}^{a} e^{-x^{2}} d x\right)^{2}. $$ (b) Let \(D_{a}\) be the disk \(x^{2}+y^{2} \leq a^{2}\). Use polar coordinates to evaluate \(\iint_{D_{a}} e^{-x^{2}-y^{2}} d x d y\). (c) Note that, as \(a\) goes to \(\infty\), both \(R_{a}\) and \(D_{a}\) fill out all of \(\mathbb{R}^{2}\). It is true that both \(\lim _{a \rightarrow \infty} \iint_{R_{a}} e^{-x^{2}-y^{2}} d x d y\) and \(\lim _{a \rightarrow \infty} \iint_{D_{a}} e^{-x^{2}-y^{2}} d x d y\) exist and that they are equal. Their common value is the improper integral \(\iint_{\mathbb{R}^{2}} e^{-x^{2}-y^{2}} d x d y\). Use this information along with your answers to parts (a) and (b) to show that: $$ \int_{-\infty}^{\infty} e^{-x^{2}} d x=\sqrt{\pi}. $$

A class of enthusiastic multivariable calculus students celebrates the change of variables theorem by drilling a cylindrical hole of radius 1 straight through the center of the earth. Assuming that the earth is a three- dimensional closed ball of radius \(2,\) find the volume of the portion of the earth that remains.

Let \(W\) be the region in \(\mathbb{R}^{3}\) lying above the \(x y\) -plane, inside the cylinder \(x^{2}+y^{2}=1\), and below the plane \(x+y+z=1\). Find the volume of \(W\).

Find the volume of the five-dimensional closed unit ball: $$ W=\left\\{\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right) \in \mathbb{R}^{5}: x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2} \leq 1\right\\}. $$ (Hint: Organize the calculation as a double integral of a triple integral: \(\left.\iint\left(\iiint \ldots\right) .\right)\)

A region \(D\) in the \(x y\) -plane is called symmetric in the line \(y=x\) if, whenever a point \((x, y)\) is in \(D,\) so is its reflection \((y, x) .\) If \(D\) is such a region, prove that its centroid lies on the line \(y=x\).
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