/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Die Halbkugel \(B=\left\\{x \in ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 25: Problem 18

Die Halbkugel \(B=\left\\{x \in \mathbb{R}^{3} \mid\|x\|0\right\\}\) besteht aus einem Material mit der Dichte \(\varrho(x)=a x_{3}, a>0\). Berechnen Sie die Masse und die dritte Koordinate des Schwerpunkts der Halbkugel.

Short Answer

Expert verified
Question: Find the total mass and the third coordinate of the center of mass of a solid hemisphere with radius R and density function 蟻(x) = ax鈧�. Answer: The mass of the hemisphere is M = (4/15)蟺aR^5, and the third coordinate of the center of mass is z鈧� = (3/8)R.

Step by step solution

01

Calculation of total mass

We first need to find the total mass of the hemisphere. We can do this by calculating the mass of an infinitesimally small element, \(dm\), in the hemisphere and then integrating over the whole hemisphere volume. In spherical coordinates, \(dm = \varrho(x) dV\), where \(\varrho(x) = a x_3 = \rho^2\sin \phi \cos \phi\) a>0 and \(dV = \rho^2 \sin \phi d\rho d\theta d\phi\). So, $$ dm = (\rho^2 \sin \phi \cos \phi)(\rho^2 \sin \phi d\rho d\theta d\phi) $$ Now, integrate over the entire hemisphere volume: $$ M = \int_0^R \int_0^{2\pi} \int_0^{\pi/2} a\rho^4 \sin^2 \phi \cos \phi d\rho d\theta d\phi $$
02

Total mass integral calculation

To find the total mass M, calculate the above triple integral. Break it into three single integrals to calculate them separately: $$ M_\rho = \int_0^R a\rho^4 d\rho = \frac{aR^5}{5} $$ $$ M_\theta = \int_0^{2\pi} d\theta = 2\pi $$ $$ M_\phi = \int_0^{\pi/2} \sin^2 \phi \cos \phi d\phi = \frac{2}{3} $$ Now, multiply the results from single integrals: $$ M = M_\rho M_\theta M_\phi = (\frac{aR^5}{5})(2\pi)(\frac{2}{3}) = \frac{4}{15}\pi aR^5 $$
03

Calculation of the third coordinate of the center of mass

To find the third coordinate of the center of mass, \(z_{cm}\), divide the moment of mass about the z-axis by the total mass: $$ z_{cm} = \frac{1}{M}\int_0^R \int_0^{2\pi} \int_0^{\pi/2} z \cdot a\rho^4 \sin^2 \phi \cos \phi d\rho d\theta d\phi $$ In spherical coordinates, z is represented as \(\rho \cos \phi\). So, $$ z_{cm} = \frac{15}{4\pi aR^5} \int_0^R \int_0^{2\pi} \int_0^{\pi/2} \rho^5 \sin^2 \phi \cos^2 \phi d\rho d\theta d\phi $$
04

Center of mass integral calculation

To find \(z_{cm}\), calculate the above triple integral. Break it into three single integrals to find them separately: $$ z_\rho = \int_0^R \rho^5 d\rho = \frac{R^6}{6} $$ $$ z_\theta = \int_0^{2\pi} d\theta = 2\pi $$ $$ z_\phi = \int_0^{\pi/2} \sin^2 \phi \cos^2 \phi d\phi = \frac{1}{3} $$ Now, multiply the results from single integrals: $$ z_{cm} = \frac{15}{4\pi aR^5} (\frac{R^6}{6})(2\pi)(\frac{1}{3})\implies = \frac{3}{8}R $$ So, the mass of the hemisphere is $$ M = \frac{4}{15}\pi aR^5 $$ and the third coordinate of the center of mass is $$ z_{cm} = \frac{3}{8}R $$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Gegeben ist die Kugelschale \(D\) um den Nullpunkt mit 盲u脽erem Radius \(R\) und innerem Radius \(r(r

Gegeben ist \(D=\left\\{x \in \mathbb{R}^{2} \mid x_{1}^{2}+x_{2}^{2}<1\right\\} .\) Berechnen Sie $$ \int_{D}\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right) \mathrm{e}^{-\left(x_{1}^{2}+x_{2}^{2}\right)} \mathrm{d} \boldsymbol{x} $$ durch Transformation auf Polarkoordinaten.

Zeigen Sie f眉r beliebige \(n \in \mathbb{N}\) die Beziehung $$ V_{n}=\int_{0}^{1} \int_{0}^{t_{1}} \ldots \int_{0}^{t_{n-1}} \mathrm{~d} t_{n} \cdots \mathrm{d} t_{2} \mathrm{~d} t_{1}=\frac{1}{n !} $$

Gegeben ist das Gebiet \(D \subseteq \mathbb{R}^{3}\), das als Schnitt der Einheitskugel mit der Menge \(\left\\{x \in \mathbb{R}^{3} \mid x_{1}, x_{2}, x_{3}>0\right\\}\) entsteht. Beschreiben Sie dieses Gebiet in kartesischen Koordinaten, Zylinderkoordinaten und Kugelkoordinaten.

Berechnen Sie die folgenden Gebietsintegrale: (a) \(J=\int_{D} \frac{\sin \left(x_{1}+x_{3}\right)}{x_{2}+2} \mathrm{~d} x\) mit \(D=\left[-\frac{\pi}{4}, 0\right] \times[0,2] \times\left[0, \frac{\pi}{2}\right]\) (b) \(J=\int_{D} \frac{2 x_{1} x_{3}}{\left(x_{1}^{2}+x_{2}^{2}\right)^{2}} \mathrm{~d} x\) mit \(D=\left[\frac{1}{\sqrt{3}}, 1\right] \times[0,1] \times[0,1]\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.