/*! 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 35 Find the coordinates of the cent... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 35

Find the coordinates of the center of mass of the following solids with variable density. The solid bounded by the upper half of the sphere \(\rho=6\) and \(z=0\) with density \(f(\rho, \varphi, \theta)=1+\rho / 4\)

Short Answer

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Question: Given a solid with a density function \(f(\rho, \varphi, \theta) = 1 + \rho / 4\), bounded by the upper half of the sphere with \(\rho = 6\) and the plane \(z = 0\), find the center of mass of the solid. Solution: To find the center of mass for this solid, we convert the spherical coordinates to Cartesian coordinates with \(f(x, y, z) = 1 + \frac{\sqrt{x^2 + y^2 + z^2}}{4}\). We calculate the total mass (M) by integrating the density function over the entire volume, then find the center of mass coordinates \((\bar{x}, \bar{y}, \bar{z})\) using the appropriate integrals. Finally, we convert the center of mass coordinates back into Cartesian coordinates.

Step by step solution

01

Convert Spherical Coordinates to Cartesian Coordinates

To find the Cartesian coordinates, we'll use the following conversion formulas: - \(x = \rho \sin\varphi \cos\theta\) - \(y = \rho \sin\varphi \sin\theta\) - \(z = \rho \cos\varphi\) For the density function, \(f(\rho, \varphi, \theta)=1+\rho / 4\), we'll first find the Cartesian equivalents of \(\rho\) and \(\varphi\) as the function only depends on these two variables. \(\rho = \sqrt{x^2 + y^2 + z^2}\) \(\cos\varphi = \frac{z}{\rho}\) Now, substitute these expressions for the density function: \(f(x, y, z) = 1 + \frac{\sqrt{x^2 + y^2 + z^2}}{4}\)
02

Compute the Total Mass

To calculate the total mass (M) of the solid, we'll integrate the density function over the whole volume: \(M = \int\int\int_V f(x, y, z) dV\) The bounds of integration are given by the upper half of the sphere, \(\rho=6\), and the plane \(z=0\): \(\rho: 0 \leq \rho \leq 6\) \(\varphi: 0 \leq \varphi \leq \pi/2\) \(\theta: 0 \leq \theta \leq 2\pi\) Using the spherical coordinates, the volume differential is given by \(dV = \rho^2 \sin\varphi d\rho d\varphi d\theta\). Integrate the density function over the volume: \(M = \int_{0}^{6} \int_{0}^{\pi/2} \int_{0}^{2\pi} (1 + \frac{\rho}{4}) \rho^2 \sin\varphi d\rho d\varphi d\theta\)
03

Find the Center of Mass

The center of mass coordinates \((\bar{x}, \bar{y}, \bar{z})\) are given by the following integrals: \(\bar{x} = \frac{1}{M} \int\int\int_V x f(x, y, z) dV\) \(\bar{y} = \frac{1}{M} \int\int\int_V y f(x, y, z) dV\) \(\bar{z} = \frac{1}{M} \int\int\int_V z f(x, y, z) dV\) We'll calculate these three integrals separately, then divide by the total mass \(M\) we computed in Step 2, and convert the resulting \(\bar{\rho},\bar{\varphi},\bar{\theta}\) to cartesian coordinates.
04

Final Computation of Center of Mass

After performing the integrations in Step 3, we'll get the coordinates of the center of mass in spherical coordinates, \(\bar{\rho}, \bar{\varphi}, \bar{\theta}\). Finally, we'll convert the center of mass coordinates back into Cartesian coordinates using the conversion formulas: - \(\bar{x} = \bar{\rho} \sin\bar{\varphi} \cos\bar{\theta}\) - \(\bar{y} = \bar{\rho} \sin\bar{\varphi} \sin\bar{\theta}\) - \(\bar{z} = \bar{\rho} \cos\bar{\varphi}\) These coordinates give us the center of mass for the solid with the given density function.

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