Problem 35 - Calculus Early Transcendentals

Question

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 \(f(\rho, \varphi, \theta)=1+\rho / 4\)

Step-by-Step Solution

Answer

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 1: 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: \(f(x, y, z) = 1 + \frac{\sqrt{x^2 + y^2 + z^2}}{4}\)

Step 2: 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: \(M = \int_{0}^{6} \int_{0}^{\pi/2} \int_{0}^{2\pi} (1 + \frac{\rho}{4}) \rho^2 \sin\varphi d\rho d\varphi d\theta\)

Step 3: 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.

Step 4: 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.