Question

Mass and center of mass Let \(S\) be a surface that represents a thin shell with density \(\rho .\) The moments about the coordinate planes (see Section 16.6 ) are \(M_{y z}=\iint_{S} x \rho(x, y, z) d S, M_{x z}=\iint_{S} y \rho(x, y, z) d S\) and \(M_{x y}=\iint_{S} z \rho(x, y, z) d S .\) The coordinates of the center of mass of the shell are \(\bar{x}=\frac{M_{y z}}{m}, \bar{y}=\frac{M_{x z}}{m},\) and \(\bar{z}=\frac{M_{x y}}{m},\) where \(m\) is the mass of the shell. Find the mass and center of mass of the following shells. Use symmetry whenever possible. The constant-density hemispherical shell \(x^{2}+y^{2}+z^{2}=a^{2}\) \(z \geq 0\)

Short answer

Question: Determine the mass and center of mass for a constant-density hemispherical shell with equation \(x^2 + y^2 + z^2 = a^2\) and \(z \geq 0\). Answer: The mass of the hemispherical shell is \(2\pi a^2 \rho\), and its center of mass is located at \((0, 0, \frac{a}{3})\).

Step-by-step solution

Find Mass (m) of the hemispherical shell

To calculate mass (m) of the hemisphere, we integrate the density over the surface: $$ m = \rho \iint_S dS $$ To perform this integration, we use polar coordinates and parameterize the surface: $$ x = a \sin\phi \cos\theta \\ y = a \sin\phi \sin\theta \\ z = a \cos\phi $$ Where \(\phi\) ranges from \(0\) to \(\frac{\pi}{2}\), and \(\theta\) ranges from \(0\) to \(2\pi\) Now, we also find the Jacobian: $$ \begin{vmatrix} \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial \theta} \\ \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial \theta} \\ \end{vmatrix} = \begin{vmatrix} a \cos\phi \cos\theta & -a \sin\phi \sin\theta \\ a \cos\phi \sin\theta & a \sin\phi \cos\theta \\ -a \sin\phi & 0 \\ \end{vmatrix} = a^2 \sin\phi $$ So the mass of the shell is: $$ m = \rho \iint_S a^2 \sin\phi d\phi d\theta $$ Integrating, we get: $$ m = \rho \int_0^{2\pi} d\theta\int_0^{\frac{\pi}{2}} a^2 \sin\phi d\phi $$ Solving this integral, we find: $$ m = 2\pi a^2 \rho $$

Find moments about the coordinate planes

Now, we find the moments about the coordinate planes using symmetry. Since the density is constant, and the hemisphere is symmetric about the xz and yz planes, the integral of x and y over the surface will be zero: $$ M_{yz} = M_{xz} = 0 $$ Now, we find \(M_{xy}\): $$ M_{xy} = \rho \iint_S z dS $$ To evaluate this integral, substitute z (from our parameterization) and Jacobian into the integral: $$ M_{xy} = \rho \int_0^{2\pi} d\theta\int_0^{\frac{\pi}{2}} a \cos\phi \cdot a^2 \sin\phi d\phi $$ Solving, we obtain: $$ M_{xy} = \frac{2\pi a^3 \rho}{3} $$

Find the center of mass

Now that we have the moments, we can find the center of mass coordinates using the given formulas: $$ \bar{x} = \frac{M_{yz}}{m} = 0 \\ \bar{y} = \frac{M_{xz}}{m} = 0 \\ \bar{z} = \frac{M_{xy}}{m} = \frac{a}{3} $$ Therefore, the center of mass for the given hemispherical shell is located at \((0, 0, \frac{a}{3})\), and its mass is \(2\pi a^2 \rho\).

Key concepts

Surface Integrals

Surface integrals allow us to calculate a quantity over a curved surface. This is similar to a double integral, but instead of integrating over a region in the plane, we integrate over the surface of a 3D shape.
To compute these integrals, we generally parameterize the surface. This means writing the surface coordinates \(x, y, z\) in terms of two parameters, usually denoted \(u\) and \(v\). For example, with a hemispherical shell, we might choose spherical coordinates where \(x = a \sin\phi \cos\theta\), \(y = a \sin\phi \sin\theta\), and \(z = a \cos\phi\).
The next step involves finding the surface element \(dS\), which is the infinitesimally small piece of the surface over which the integral is conducted. For spherical coordinates, this can be calculated using the Jacobian determinant, offering a key role in the transformation, given as \(a^2 \sin\phi\).
Using surface integrals, one can calculate properties such as mass, by integrating the density over the whole surface, as demonstrated in the hemispherical shell problem.

Coordinate Systems

To simplify surface integrals, different coordinate systems are brought into play. The choice of the correct coordinate system can simplify calculations significantly, especially when dealing with symmetrical objects.

• **Cartesian Coordinates**: Composed of \(x, y, z\) and effective for rectangular, cubical spaces.
• **Polar Coordinates**: Utilize \(r\) and \(\theta\) for circular or cylindrical symmetry.
• **Spherical Coordinates**: Best for spheres or hemispheres, expressed as \(\phi\) (polar angle) and \(\theta\) (azimuthal angle), while \(\rho\) remains constant for a surface.

In the hemispherical shell exercise, spherical coordinates are chosen because they align well with the symmetry of the problem. This makes the resulting equations easier to handle, as they naturally fit the shape of the hemisphere.
Overall, picking the suitable coordinate system can turn a complex task into a much simpler problem, leveraging the object's symmetry.

Moment of Inertia

The moment of inertia measures how a body's mass is distributed relative to an axis and affects how the object rotates. For surfaces, this is calculated by integrating the square of the distance each element of mass is from the axis, multiplied by the surface density.
In our exercise, the moments about the coordinate planes \(M_{yz}, M_{xz},\) and \(M_{xy}\) are analogous to inertia. Here, they take into account the surface density and symmetry when calculating the mass distribution.
The symmetry of the hemispherical shell simplifies many calculations. For instance, because the hemisphere is symmetric around the \(xz\) and \(yz\) planes, we find \(M_{yz} = M_{xz} = 0\). This means the mass distribution doesn't cause any rotation about these axes.
Finally, once we have these moments, they lend insight into dynamics by potentially guiding behavior like balance or rotation. It's an integral part of both structural and mechanical engineering analysis.