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Chapter 13: Problem 56

Consider the following two-and three-dimensional regions. Specify the surfaces and curves that bound the region, choose a convenient coordinate system, and compute the center of mass assuming constant density. All parameters are positive real numbers. A solid is enclosed by a hemisphere of radius \(a\). How far from the base is the center of mass?

Short Answer

Expert verified
Answer: The center of mass of the solid hemisphere is located at a distance of \(\frac{3}{4}a\) units away from the base.

Step by step solution

01

Identify the bounded region and coordinate system

The region is enclosed by a hemisphere of radius \(a\) and its base is a circular disk. The coordinate system that is most convenient for this problem is the spherical coordinate system, with the origin at the center of the base. In spherical coordinates, the hemisphere can be described as follows: - The radial distance, \(r\), is between \(0\) and \(a\) - The polar angle, \(\theta\), is between \(0\) and \(\frac{\pi}{2}\) - The azimuthal angle, \(\phi\), goes all around the base with \(0\) to \(2\pi\).
02

Set up the mass element

To find the center of mass, we need to use the density and the mass element, \(dm\). The mass element can be described in spherical coordinates as $$dm = \rho dV= \rho r^2 \sin\theta dr d\theta d\phi$$ where \(\rho\) is the constant density.
03

Calculate the total mass

We first calculate the total mass of the hemisphere, which can be obtained by integrating the mass element, \(dm\), over the volume of the hemisphere: $$M = \int dm = \rho \int_{0}^{a}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi} r^2 \sin\theta dr d\theta d\phi$$ To calculate the total mass, we compute the integral: $$M = \rho \int_{0}^{a}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi} r^2 \sin\theta dr d\theta d\phi = \frac{2}{3}\pi a^3 \rho$$
04

Calculate the moments

To find the center of mass, we need to compute the moments of the mass distribution with respect to each axis. Since we are looking for the distance from the base of the hemisphere, we only need to calculate the moment around the base (the z-axis). We do this by integrating the z-component of the position vector, which is \(r\cos \theta\), with respect to the mass element: $$M_z = \int z dm = \rho \int_0^a\int_0^\frac{\pi}{2}\int_0^{2\pi} r^3\cos\theta\sin\theta dr d\theta d\phi$$ Compute the integral to find the moment: $$M_z = \rho \int_0^a\int_0^\frac{\pi}{2}\int_0^{2\pi} r^3\cos\theta\sin\theta dr d\theta d\phi = \frac{1}{2}\pi a^4 \rho$$
05

Compute the center of mass

Divide the moment \(M_z\) by the total mass \(M\) to obtain the z-coordinate of the center of mass, which is the distance from the base of the hemisphere: $$z_{cm} = \frac{M_z}{M} = \frac{\frac{1}{2}\pi a^4 \rho}{\frac{2}{3}\pi a^3 \rho} = \frac{3}{4}a$$ Thus, the center of mass of the solid hemisphere is \(\frac{3}{4}a\) units away from the base.

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