Question

Use spherical polar coordinates \(r, \theta, \phi\) to find the CM of a uniform solid hemisphere of radius R, whose flat face lies in the \(x y\) plane with its center at the origin. Before you do this, you will need to convince yourself that the element of volume in spherical polars is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

Short answer

The center of mass is at \((0, 0, \frac{3R}{8})\).

Step-by-step solution

Understand the Volume Element

In spherical coordinates, the differential volume element is given by \(dV = r^2 \sin \theta \, dr \, d\theta \, d\phi\). This accounts for the stretching of coordinates as they move away from the origin.

Set Up the Integral for Mass

Since the hemisphere is uniform, the mass is proportional to the volume. The total mass \(M\) is the integral of the density, \(\rho\), over the volume. Here, \(\rho = \text{constant}\), so we integrate over the hemisphere:\[M = \rho \int_0^{R} \int_0^{\pi/2} \int_0^{2\pi} r^2 \sin \theta \, dr \, d\theta \, d\phi\]

Calculate the Mass

Perform the integration to find the total mass \(M\). The angular integrals are:For \(\phi\):\[\int_0^{2\pi} d\phi = 2\pi\]For \(\theta\):\[\int_0^{\pi/2} \sin \theta \, d\theta = 1\]And finally integrate over \(r\):\[\int_0^R r^2 \, dr = \frac{R^3}{3}\]Thus, the total mass is:\[M = \rho \cdot 2\pi \cdot 1 \cdot \frac{R^3}{3} = \frac{2\pi \rho R^3}{3}\]

Find the Center of Mass in Spherical Coordinates

The center of mass (CM) has coordinates \((\bar{r}, \bar{\theta}, \bar{\phi})\), but since we expect\y-axis symmetry and translational symmetry in \(\phi\), the CM should have \(\bar{\phi} = 0\) and \(\bar{r} = 0\). Using symmetric arguments and isotropy along the \(\phi\) direction, the relevant calculation remains for \(z\)-coordinate (vertical). Convert \((r, \theta, \phi)\) into Cartesian coordinates. For the \(z\)-coordinate:\[z = r \cos \theta\]

Calculate the Center of Mass Coordinate

The \(z\)-coordinate of the CM is calculated as:\[\bar{z} = \frac{1}{M} \int_0^R \int_0^{\pi/2} \int_0^{2\pi} (r \cos \theta) r^2 \sin \theta \, dr \, d\theta \, d\phi\]Breaking this down includes the angular and radial integration:- \[ \int_0^{2\pi} d\phi = 2\pi \]- \[ \int_0^{\pi/2} \sin \theta \cos \theta \, d\theta = \left[ \frac{\sin^2 \theta}{2} \right]_0^{\pi/2} = \frac{1}{2} \]- \[ \int_0^R r^3 \, dr = \frac{R^4}{4} \]Thus:\[\bar{z} = \frac{1}{\frac{2\pi \rho R^3}{3}} \cdot 2\pi \cdot \frac{1}{2} \cdot \frac{R^4}{4} \]Solving:\[\bar{z} = \frac{3R}{8}\]

Summarize the Coordinates of the Center of Mass

The center of mass is at \((0, 0, \frac{3R}{8})\) since \(\bar{x} = 0\) and \(\bar{y} = 0\) due to symmetry, and the calculated \(\bar{z}\) as above.

Key concepts

Spherical Coordinates

When dealing with three-dimensional problems, spherical coordinates offer a convenient way to express points in space. Unlike Cartesian coordinates, which use \((x, y, z)\), spherical coordinates are described by three parameters: \(r\), \(\theta\), and \(\phi\).

- \(r\) denotes the radial distance from the origin, extending outwards. - \(\theta\) is the polar angle measured down from the positive \(z\)-axis. - \(\phi\) is the azimuthal angle around the \(z\)-axis, similar to longitude.

This system of coordinates is particularly useful for problems involving spheres or spherical symmetry, such as finding the center of mass of a hemisphere. By employing these coordinates, one can easily account for changes in dimensions that occur as you move further from the origin, particularly when integrating over volumes.

Hemisphere

A hemisphere is half of a sphere, resembling a solid dome. Given its shape, it has rotational symmetry around its central vertical axis, which simplifies many calculations. In this problem, we focus on a uniform solid hemisphere with its flat face lying on the \(xy\) plane and centered at the origin. This configuration implies:

- The hemisphere extends from the bottom plane upward to its curved surface. - Symmetry along the \(z\)-axis helps in reducing the complexity of mathematical computation, especially while determining the center of mass.

The center of mass for this shape must factor in its geometric symmetry, particularly because any shifts in density would affect its location. For a uniform hemisphere, the symmetrical nature around the vertical axis means that the center of mass lies on this axis.

Volume Element

In spherical coordinates, the volume element \(dV\) is not simply \(dr\,d\theta\,d\phi\). Instead, it takes the form \(dV = r^2 \sin \theta \, dr \, d\theta \, d\phi\).

This expression derives from how distances and areas stretch when moving away from the central point.

• The factor \(r^2\) represents the radial scaling as you move outward from the origin.
• \(\sin \theta\) accounts for the variation in distance along the axis at different \(\theta\) angles.

It's important to remember this when setting up integrals, as incorrect volume element expressions can lead to erroneous results. To calculate physical properties like mass or center of mass, integrating over the entire volume element correctly is crucial.

Integration in Polar Coordinates

Integrating in spherical coordinates involves multiple integrations across different dimensions: radial, polar, and azimuthal. The general approach is similar to using polar coordinates, with additional complexity arising from the third dimension.

Here’s how you would typically set it up for a uniform solid hemisphere:

• Start by integrating over \(\phi\), as it typically spans from 0 to \(2\pi\) and accounts for the full circle around the \(z\)-axis.
• Next, the integral over \(\theta\) goes from 0 to \(\pi/2\), reflecting the dome-like shape of the hemisphere.
• Finally, the integral over \(r\) spans from 0 to the hemisphere's radius \(R\).

The order and limits of integration are determined by the physical shape and how it is oriented in space. When properly handled, this integration helps find quantities such as mass and the precise position of the center of mass.