Question

Find the center of mass of the following solids, assuming a constant density of 1. Sketch the region and indicate the location of the centroid. Use symmetry when possible and choose a convenient coordinate system. The region bounded by the upper half \((z \geq 0)\) of the ellipsoid \(4 x^{2}+4 y^{2}+z^{2}=16\)

Short answer

Considering an ellipsoid with the equation \(4x^2 + 4y^2 + z^2 = 16\) and constant density of 1, we can use symmetry and spherical coordinates to calculate that its centroid (center of mass) is at the origin, \((0, 0, 0)\).

Step-by-step solution

Use symmetry

Since the ellipsoid is symmetric with respect to the x, y, and z axes, the centroid (center of mass) must lie on each of these axes, so \(M_x = M_y = M_z\). Symmetry allows us to find M_x and know that M_y and M_z are equal to M_x.

Choose a convenient coordinate system

To simplify the integration process, let's use the spherical coordinate system, with \(x = \rho\sin\phi\cos\theta\), \(y = \rho\sin\phi\sin\theta\), and \(z = \rho\cos\phi\). The Jacobian determinant is \(J = \rho^2\sin\phi\). Additionally, the equation of the ellipsoid in spherical coordinates is \(\rho^2 = 16\sin^2\phi\cos^2\theta + 16\sin^2\phi\sin^2\theta + 16\cos^2\phi = 16\), so \(\rho = 2\sqrt{2}\).

Calculate the volume of the ellipsoid

The volume of the ellipsoid V is given by: \(V = \iiint\limits_D \rho^2\sin\phi d\rho d\phi d\theta = \int_{0}^{2\sqrt{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\rho^2\sin\phi d\rho d\phi d\theta\) \(V = \int_{0}^{2\sqrt{2}}\rho^2 d\rho \int_{0}^{\frac{\pi}{2}}\sin\phi d\phi \int_{0}^{2\pi} d\theta\) \(V = \left[\frac{1}{3}\rho^3\right]_{0}^{2\sqrt{2}} \cdot \left[-\cos\phi\right]_{0}^{\frac{\pi}{2}} \cdot \left[\theta\right]_{0}^{2\pi}\) \(V = \frac{32}{3} \cdot 1 \cdot 2\pi = \frac{64}{3}\pi\)

Calculate \(M_x\)

Using the formula for \(M_x\), we can write: \(M_x = \frac{1}{V}\iiint\limits_D x dV = \frac{1}{V} \iiint\limits_D \rho\sin\phi\cos\theta(\rho^2\sin\phi)d\rho d\phi d\theta\) Since the bounds of the region are given in spherical coordinates and we've already found the volume V: \(M_x = \frac{3}{64\pi}\int_{0}^{2\sqrt{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi}\rho^3\sin^2\phi \cos\theta d\rho d\phi d\theta\) \(M_x = \frac{3}{64\pi}\left(\int_{0}^{2\sqrt{2}}\rho^3 d\rho\right)\left(\int_{0}^{\frac{\pi}{2}}\sin^2\phi d\phi\right)\left(\int_{0}^{2\pi}\cos\theta d\theta\right)\) Notice that the last integral equals zero, which indicates that \(M_x = 0\).

Use symmetry to find M_y and M_z

Since the ellipsoid is symmetric with respect to the x, y, and z axes, we know that \(M_y = M_z = M_x\). As we found that \(M_x = 0\), both \(M_y\) and \(M_z\) are also 0.

Sketch the region and centroid

We can now sketch the region as an ellipsoid with equation \(4x^2 + 4y^2 + z^2 = 16\), lying in the upper half-plane (\(z\geq 0\)), and indicate the location of the centroid at \((0, 0, 0)\). Due to symmetry, the centroid is located at the center of the ellipsoid.

Key concepts

Symmetric Solids

Symmetric solids, such as ellipsoids, have equal geometric properties along their axes. In the example of the ellipsoid described by the equation \(4x^2 + 4y^2 + z^2 = 16\), symmetry around the x, y, and z axes plays a crucial role in finding the center of mass. When dealing with symmetric solids, symmetry allows us to make useful assumptions. Because the ellipsoid is symmetric around all three axes, the center of mass, or centroid, lies on each axis at the coordinate origin \((0, 0, 0)\). This simplifies calculations significantly, as only one integral needs to be computed to find the mass moment (\(M_x\)), with the others being equal due to symmetry. Utilizing symmetry helps in gaining intuition about problem-solving in physics and is a powerful tool for reducing complexity during analysis.

Spherical Coordinates

Spherical coordinates provide a powerful alternative to Cartesian coordinates for analyzing three-dimensional figures, particularly those with symmetry, like spheres and ellipsoids. The key variables used in spherical coordinates are \(\rho\), \(\phi\), and \(\theta\).

• \(\rho\) represents the distance from the origin to a point.
• \(\phi\) is the polar angle, which measures the angle from the positive z-axis.
• \(\theta\) is the azimuthal angle, measured in the xy-plane from the positive x-axis.

For the ellipsoid problem, the transformation from Cartesian to spherical coordinates aids in simplifying the integration process. By expressing the ellipsoid's equation \(4x^2 + 4y^2 + z^2 = 16\) in spherical coordinates, we get \(\rho = 2\sqrt{2}\), which assists in setting up the bounds for the volume integral. Spherical coordinates not only simplify calculations but also provide insights into the geometry and symmetry of the solid being analyzed.

Ellipsoid

An ellipsoid is a three-dimensional geometric shape, akin to a stretched or compressed sphere. Its surface can be mathematically defined with an equation such as \(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\). In this exercise, the ellipsoid is sectioned into an upper half by the condition \(z \geq 0\), creating a dome-like shape. The specific form \(4x^2 + 4y^2 + z^2 = 16\) stacks the semi-axes with different scaling factors, \(2\) along x and y, and \(1\) along z, denoted as an oblate spheroid. The process for finding its center of mass leverages its symmetrical properties, and through integration, we assess how mass distributes within this solid's confines.Understanding the structure and symmetry of an ellipsoid is essential for performing integrations and applying coordinate transformations effectively.

Integration in Physics

Integration is a fundamental technique in physics for computing properties of continuous bodies. When finding properties like volume or center of mass, integration allows for the summation of infinitesimal elements over a given region. In the context of the ellipsoid, integration helps determine the volume and the centers of mass along the x, y, and z axes, using limits defined by spherical coordinates. The integral:\[ V = \int_{0}^{2\sqrt{2}}\rho^2 d\rho \int_{0}^{\frac{\pi}{2}}\sin\phi d\phi \int_{0}^{2\pi} d\theta \]calculates the volume of the half-ellipsoid.For the center of mass, the integration yields zero for \(M_x\), \(M_y\), and \(M_z\) because the cosine function, representing the projection on the respective axes, averages to zero over its symmetric range. Understanding these principles of integration in physics helps in tackling complex spatial problems and is essential for accurately analyzing symmetric solids like the ellipsoid.