Question

A solid is described along with its density function. Find the center of mass of the solid using cylindrical coordinates. (Note: these are the same solids and density functions as found in Exercises 23 through 26.) The upper half of the unit ball, bounded between \(z=0\) and \(z=\sqrt{1-x^{2}-y^{2}},\) with density function \(\delta(x, y, z)=1\).

Short answer

Center of mass is at \((0, 0, \frac{\pi}{4})\).

Step-by-step solution

Set Up the Problem

We need to find the center of mass of the solid, which is the upper half of the unit ball. The density function is constant, \( \delta(x, y, z) = 1 \). Because we are working with a spherical shape, cylindrical coordinates make the problem easier. In cylindrical coordinates, \((x, y, z)\) convert to \((r, \theta, z)\) with \(x = r \cos \theta\), \(y = r \sin \theta\), and \(z = z\). The bounds for the upper half unit ball are \(0 \leq r \leq 1\), \(0 \leq \theta \leq 2\pi\), and \(0 \leq z \leq \sqrt{1 - r^2}\).

Formula for Center of Mass

The center of mass \((\bar{x}, \bar{y}, \bar{z})\) is found using:\[\bar{x} = \frac{1}{M} \int \int \int x \delta(x, y, z) \, dV, \\bar{y} = \frac{1}{M} \int \int \int y \delta(x, y, z) \, dV, \\bar{z} = \frac{1}{M} \int \int \int z \delta(x, y, z) \, dV\]where \( M \) is the total mass, given by \( M = \int \int \int \delta(x, y, z) \, dV \). Because \(\delta(x, y, z) = 1\), these formulas simplify to the integrals across the solid V.

Calculate Total Mass \(M\)

Using the formula for mass with \( \delta(x, y, z) = 1 \), we integrate:\[M = \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{\sqrt{1-r^2}} r \, dz \, dr \, d\theta.\]Calculate the innermost integral: \[ \int_{0}^{\sqrt{1-r^2}} 1 \, dz = \sqrt{1-r^2}. \]The full integral is:\[M = \int_{0}^{2\pi} \int_{0}^{1} \sqrt{1-r^2} \, r \, dr \, d\theta.\]

Solve for \(\bar{x}\) and \(\bar{y}\)

Since the density function is uniform and the object is symmetric about the z-axis, \(\bar{x}\) and \(\bar{y}\) are both zero. The integrals lead to zero because for every positive \(x\) or \(y\), there is an equivalent negative in this symmetrical shape, cancelling out to zero.

Calculate \(\bar{z}\)

To find \(\bar{z}\), we compute:\[\bar{z} = \frac{1}{M} \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{\sqrt{1-r^2}} z \, r \, dz \, dr \, d\theta.\]Calculate the innermost integral: \[ \int_{0}^{\sqrt{1-r^2}} z \, dz = \left[ \frac{z^2}{2} \right]_{0}^{\sqrt{1-r^2}} = \frac{1-r^2}{2}.\]The integral becomes:\[\int_{0}^{2\pi} \int_{0}^{1} \frac{1-r^2}{2} \, r \, dr \, d\theta.\]

Solve Completely

Calculate the \(\bar{z}\) integral by evaluating:\[\int_{0}^{1} \frac{r(1-r^2)}{2} \, dr = \left[ \frac{r^2}{4} - \frac{r^4}{8} \right]_{0}^{1} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}.\]Then multiply by \(2\pi\):\[ 2\pi \times \frac{1}{8} = \frac{\pi}{4}.\]Finally, the center of mass is \((0, 0, \frac{\pi}{4})\).

Conclusion

The center of mass for the upper half of the unit ball with the given density function is calculated to be \((0, 0, \frac{\pi}{4})\).

Key concepts

Cylindrical Coordinates

Cylindrical coordinates make it easier to handle problems involving circular symmetry. They are a system of coordinates that are essentially a blend of polar coordinates and Cartesian coordinates. The major difference compared to Cartesian coordinates is the use of radius and angle to describe positions in space.
In cylindrical coordinates, a point in three-dimensional space is represented by

• The radial distance from the z-axis, denoted by \(r\). It represents how far the point is from the z-axis.
• The angle \(\theta\) in the \((x, y)\) plane, which indicates the direction from the positive x-axis.
• The height along the z-axis, denoted by \(z\).

Using these coordinates can significantly simplify the process of calculating integrals over symmetric shapes, such as cylinders or spheres. In our problem, transforming the Cartesian coordinates \((x, y, z)\) to cylindrical coordinates \((r, \theta, z)\) facilitates working with the upper half of the unit ball by directly applying bounds suitable for such symmetry.

Density Function

A density function describes how mass is distributed throughout a region. It can differ based on the type of material or properties of the problem at hand. In our exercise, however, the density function is kept constant at \(\delta(x, y, z) = 1\).
This simplification states that the mass is evenly distributed across the volume of the object. Understandably, a constant density results in a simplified computation because it allows the mass integrals to focus solely on the volume of the solid to find total mass and center of mass.
The straightforward nature of this density function is due to both the simplicity of the given system and the solid’s uniformity, making it an insightful first step for learning about such calculations. It enables students to focus more on understanding the geometry and physics behind center of mass calculations without additional complexity introduced by variable densities.

Cylindrical Integration

Cylindrical integration is used when dealing with volumes and shapes that possess rotational symmetry around an axis, commonly the z-axis in cylindrical coordinates.
Because cylindrical coordinates inherently suit such symmetric objects, the integration bounds match those coordinates:

• The radial direction \(r\) typically ranges from the axis outward (e.g., from 0 to 1 for unit shapes).
• The angular direction \(\theta\) commonly rotates full circle \(0\) to \(2\pi\).
• The vertical height \(z\) varies according to the shape’s constraint, such as \(0\) to \(\sqrt{1-r^2}\) for the upper half of a sphere.

By converting the integration to a volume element \(dV = r \, dz \, dr \, d\theta\), each integral smoothly translates volume changes into changes in mass. This solidifies the link between shape geometry and mathematical computation, enabling precise volume and mass determination.

Symmetric Solid

A symmetric solid is a shape that is balanced in such a way that equal subdivisions of the solid mirror each other. In three dimensions, this often implies rotational or reflective symmetry. In the context of our problem, we consider the upper half of the unit sphere, which is perfectly symmetric about the z-axis.
This symmetry is crucial because it simplifies the calculations for the center of mass.

• For symmetry about the z-axis, similar mass distributions exist on either side of the z coordinate, meaning the x and y coordinates for the center of mass automatically balance out to zero.
• The mass is uniformly distributed, thanks to the constant density function, accentuating this symmetry.
• Thus, focus shifts to calculating \(\bar{z}\), the vertical shift of the center of mass within the given bounds, where mass distribution along z (but not x or y) is non-symmetrical due to the semi-spherical shape.

Recognizing and leveraging the symmetrical nature of solids plays a vital role in simplifying and correctly solving integration problems in physics and calculus.