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Chapter 14: 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

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Question: Determine the location of the center of mass of a solid hemisphere with radius \(a\) and constant density. Answer: The center of mass of the solid hemisphere is located at \((0, 0, \frac{3}{8}a)\) and is \(\frac{3}{8}a\) away from the base.

Step by step solution

01

Identify the bounding surfaces and curves

For a solid hemisphere of radius \(a\), the bounding surfaces and curves include: 1. The curved surface of the hemisphere. 2. The circular base of radius \(a\) on the xy-plane.
02

Choose a convenient coordinate system

Considering the symmetry of the hemisphere, we will choose the spherical coordinate system as it is most suitable for this problem. Spherical coordinates use three variables: \(蟻\) (rho) representing the distance from the origin, \(胃\) (theta) as the polar angle, and \(蠁\) (phi) as the azimuthal angle. The coordinate system transformation from spherical coordinates to Cartesian coordinates is given as: $$x = 蟻sin(胃)cos(蠁)$$ $$y = 蟻sin(胃)sin(蠁)$$ $$z = 蟻cos(胃)$$
03

Compute the center of mass

As the density is constant, we can compute the center of mass using mass-weighted integrals divided by the total mass of the hemisphere. Using spherical coordinates, the volume element is given by: \(dV = 蟻^2sin(胃)d蟻d胃d蠁\). Let's denote the constant density as 蟻\(_0\). Then, the total mass of the hemisphere can be calculated as: $$M = 蟻_0 \int_0^a\int_0^蟺\int_0^{2蟺} 蟻^2sin(胃)d蟻d胃d蠁 = 蟻_0 \left[\frac{1}{3}蟻^3\right]_0^a \left[-cos(胃)\right]_0^蟺 \left[蠁\right]_0^{2蟺} = \frac{2}{3}蟺a^3蟻_0$$ The center of mass components can be computed as: $$x_c = \frac{1}{M} \int_0^a\int_0^蟺\int_0^{2蟺} x 蟻^2sin(胃)d蟻d胃d蠁 = \frac{1}{M} \int_0^a\int_0^蟺\int_0^{2蟺} 蟻sin^2(胃)cos(蠁)d蟻d胃d蠁$$ $$y_c = \frac{1}{M} \int_0^a\int_0^蟺\int_0^{2蟺} y 蟻^2sin(胃)d蟻d胃d蠁 = \frac{1}{M} \int_0^a\int_0^蟺\int_0^{2蟺} 蟻sin^2(胃)sin(蠁)d蟻d胃d蠁$$ As the hemisphere is symmetrical about the z-axis, the x and y components of the center of mass will be zero. Therefore, we only need to compute the z-component of the center of mass: $$z_c = \frac{1}{M} \int_0^a\int_0^蟺\int_0^{2蟺} z 蟻^2sin(胃)d蟻d胃d蠁 = \frac{1}{M} \int_0^a\int_0^蟺\int_0^{2蟺} 蟻^3cos(胃)sin(胃)d蟻d胃d蠁$$ Performing the integration, we get: $$z_c = \frac{1}{\frac{2}{3}蟺a^3蟻_0} \left[\frac{1}{4}蟻^4\right]_0^a \left[sin^2(胃)\right]_0^蟺 \left[蠁\right]_0^{2蟺} = \frac{3}{8}a$$ Thus, the center of mass of the solid hemisphere is located at \((0, 0, \frac{3}{8}a)\) and is \(\frac{3}{8}a\) away from the base.

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Key Concepts

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

Center of Mass
The center of mass is a critical concept in physics and engineering, which helps determine the average position of all mass in a system. For any solid object, regardless of its shape, the center of mass is the point at which it can be balanced perfectly if supported. Consider it as the "average" location of the object's mass.
The calculation of the center of mass depends on symmetry and density. When we have symmetrical objects like spheres or hemispheres, we often find that the center of mass lies along the axis of symmetry, which simplifies calculations. In this problem with a hemisphere, because the symmetry aligns along the azimuthal angle, the x and y components are zero, focusing only on the z-axis.
This allows us to compute the center of mass by integrating over the volume of the object and dividing by the object's total mass. The formula for the center of mass in this case simplifies the process due to these symmetrical properties.
Spherical Coordinates
In problems with spherical symmetry, using spherical coordinates often simplifies the mathematics. Unlike Cartesian coordinates, which make calculations cumbersome in such problems, spherical coordinates take advantage of symmetry.
In this system:
  • \(蟻\) indicates the radial distance from the origin (similar to the radius in polar coordinates).
  • \(胃\) (theta) is known as the polar angle, measured from the positive z-axis.
  • \(蠁\) (phi) is the azimuthal angle, loosely equivalent to the angle in polar coordinates on the xy-plane.
These three parameters help describe any point in three-dimensional spaces, especially ones with radial symmetry like spheres or hemispheres, more naturally and compactly. When evaluating integrals, the volume element in spherical coordinates is expressed as \(dV = 蟻^2sin(胃)d蟻d胃d蠁\). This expression allows us to replace Cartesian differentials with those fitting the symmetry, simplifying solving the problem.
Integration
Integration is a process of calculating the area or volume that encompasses the sum of all its infinitesimal parts. In calculus, this process becomes vital for finding the center of mass when you need to consider how to account for every particle in a volume.
To compute the center of mass in three-dimensional objects like hemispheres, triple integrals are employed. Here's how:
  • **Mass Calculation**: First, you need to calculate total mass integrating over the whole volume, usually using a density multiplied by the differential volume element, which in this problem is \(蟻^2sin(胃)d蟻d胃d蠁\).
  • **Center of Mass Components**: Next, you find individual center of mass components for the x, y, and z axes, though symmetry can sometimes simplify this. Each is found through similar integrals that multiply differentials by their respective coordinates.
  • **Normalization**: Finally, each center of mass component is divided by the total mass, ensuring the result represents the average position.
These steps outline the straightforward yet methodical approach necessary to integrate over complicated geometry, like a hemisphere, contributing to a bigger solution.
Symmetry in Geometry
Symmetry simplifies complex problems, particularly in geometry and calculus. By using symmetry, we reduce the burden of computation and focus on key features like axes of symmetry.
In the case of the solid hemisphere, symmetry is present about the z-axis. This symmetry implies that for any point at \((x, y, z)\), there exists a corresponding point \((-x, -y, z)\). Because of this symmetry in x and y directions, the respective center of mass components cancel out, simplifying to a component purely along the z-axis.
Understanding symmetry can guide you to select the most appropriate coordinate system, such as spherical coordinates in this hemisphere problem, ensuring calculations better accommodate the object's shape. This concept keeps us focused on less complicated paths to solutions as it lets us discard unnecessary computations.

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Most popular questions from this chapter

A thin plate of unit density occupies the region between the parabola \(y=a x^{2}\) and the horizontal line \(y=b\) where \(a > 0\) and \(b > 0 .\) Show that the center of mass is \(\left(0, \frac{3 b}{5}\right),\) independent of \(a\)

Evaluate the following integrals. $$\iint_{R} x \sec ^{2} y d A ; R=\left\\{(x, y): 0 \leq y \leq x^{2}, 0 \leq x \leq \sqrt{\pi} / 2\right\\}$$

Diamond region Consider the region \(R=\\{(x, y):|x|+|y| \leq 1\\}\) shown in the figure. a. Use a double integral to show that the area of \(R\) is 2 b. Find the volume of the square column whose base is \(R\) and whose upper surface is \(z=12-3 x-4 y\) c. Find the volume of the solid above \(R\) and beneath the cylinder \(x^{2}+z^{2}=1\) d. Find the volume of the pyramid whose base is \(R\) and whose vertex is on the \(z\) -axis at (0,0,6)

The Jacobian is a magnification (or reduction) factor that relates the area of a small region near the point \((u, v)\) to the area of the image of that region near the point \((x, y)\) a. Suppose \(S\) is a rectangle in the \(u v\) -plane with vertices \(O(0,0)\) \(P(\Delta u, 0),(\Delta u, \Delta v),\) and \(Q(0, \Delta v)\) (see figure). The image of \(S\) under the transformation \(x=g(u, v), y=h(u, v)\) is a region \(R\) in the \(x y\) -plane. Let \(O^{\prime}, P^{\prime},\) and \(Q^{\prime}\) be the images of O, P, and \(Q,\) respectively, in the \(x y\) -plane, where \(O^{\prime}\) \(P^{\prime},\) and \(Q^{\prime}\) do not all lie on the same line. Explain why the coordinates of \(\boldsymbol{O}^{\prime}, \boldsymbol{P}^{\prime}\), and \(Q^{\prime}\) are \((g(0,0), h(0,0))\) \((g(\Delta u, 0), h(\Delta u, 0)),\) and \((g(0, \Delta v), h(0, \Delta v)),\) respectively. b. Use a Taylor series in both variables to show that $$\begin{aligned} &g(\Delta u, 0) \approx g(0,0)+g_{u}(0,0) \Delta u\\\ &g(0, \Delta v) \approx g(0,0)+g_{v}(0,0) \Delta v\\\ &\begin{array}{l} h(\Delta u, 0) \approx h(0,0)+h_{u}(0,0) \Delta u \\ h(0, \Delta v) \approx h(0,0)+h_{v}(0,0) \Delta v \end{array} \end{aligned}$$ where \(g_{u}(0,0)\) is \(\frac{\partial x}{\partial u}\) evaluated at \((0,0),\) with similar meanings for \(g_{v}, h_{u}\) and \(h_{v}\) c. Consider the vectors \(\overrightarrow{O^{\prime} P^{\prime}}\) and \(\overrightarrow{O^{\prime} Q^{\prime}}\) and the parallelogram, two of whose sides are \(\overrightarrow{O^{\prime} P^{\prime}}\) and \(\overrightarrow{O^{\prime} Q^{\prime}}\). Use the cross product to show that the area of the parallelogram is approximately \(|J(u, v)| \Delta u \Delta v\) d. Explain why the ratio of the area of \(R\) to the area of \(S\) is approximately \(|J(u, v)|\)

Use a change of variables to evaluate the following integrals. \(\iiint_{D} d V ; D\) is bounded by the upper half of the ellipsoid \(x^{2} / 9+y^{2} / 4+z^{2}=1\) and the \(x y\) -plane. Use \(x=3 u\) \(y=2 v, z=w\)
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