Question

Let \(S\) be a surface that represents a thin shell with density \(\rho .\) The moments about the coordinate planes (see Section 13.6 ) 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 \(\bar{x}=\frac{M_{y z}}{m}, \bar{y}=\frac{M_{x z}}{m}, \bar{z}=\frac{M_{x y}}{m},\) 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 half cylinder \(x^{2}+z^{2}=a^{2},-h / 2 \leq y \leq h / 2, z \geq 0\)

Short answer

\(x_u = -a\sin(u)\), \(y_u = 0\), \(z_u = a\cos(u)\), \(x_v = 0\), \(y_v = 1\), \(z_v = 0\). Then, the differential of surface area can be found by computing the cross product of the partial derivative vectors and finding its magnitude: \(\frac{\partial(x,y,z)}{\partial(u,v)} =\lVert \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ -a\sin(u) & 0 & a\cos(u) \\ 0 & 1 & 0 \end{vmatrix} \rVert = \lVert \begin{vmatrix} a\cos(u) \\ 0 \\ 0 \end{vmatrix} \rVert = a\cos(u)\). Now, we can find the double integral: \(\iint_{S} dS = \int_{-h/2}^{h/2}\int_{0}^{\pi} a\cos(u) dvdu\) Integrate on \(v\): \(\int_{-h/2}^{h/2}\int_{0}^{\pi} a\cos(u) dvdu = a\cos(u) \int_{-h/2}^{h/2} dv \int_{0}^{\pi} du\) Integrate on \(u\): \(a\cos(u) \int_{-h/2}^{h/2} dv \int_{0}^{\pi} du = a\sin(u) \Big|_{0}^{\pi} \cdot h = 2ah\) Now, we can find the mass of the shell: \(m = \rho \iint_{S} dS = 2ah\rho\) #tag_title#Step 2: Calculate the center of mass#tag_content# The center of mass coordinates are given by the equations: \(\bar{x} = \frac{1}{m}\iint_{S}x dS\) \(\bar{y} = \frac{1}{m}\iint_{S}y dS\) \(\bar{z} = \frac{1}{m}\iint_{S}z dS\) We can simplify the calculations using symmetry of the half cylinder. Since the cylinder is symmetrical around the y-axis, we know that \(\bar{x} = 0\). Similarly, because the cylinder is symmetrical about \(y = 0\), we know that \(\bar{y} = 0\). Now, we just need to find \(\bar{z}\): \(\bar{z} = \frac{1}{m}\iint_{S}z dS = \frac{1}{2ah\rho}\iint_{S}z dS = \frac{1}{2ah\rho}\int_{-h/2}^{h/2}\int_{0}^{\pi} a\cos(u) \cdot a\sin(u) dvdu\) Integrate on \(v\): \(= \frac{a^2\sin(u)\cos(u)}{2ah\rho} \cdot h \int_{0}^{\pi} du\) Integrate on \(u\): \(= \frac{a^2h\sin(u)\cos(u)}{4ah\rho} \Big|_{0}^{\pi} = 0\) So, the center of mass is \(\bar{x} = 0\), \(\bar{y} = 0\), and \(\bar{z} = 0\). #tag_title#Conclusion#tag_content# The mass and center of mass of the constant-density half cylinder with the given equation \(x^2 + z^2 = a^2\) and \(-h/2 \leq y \leq h/2\), \(z\geq0\) are: Mass: \(m = 2ah\rho\) Center of mass: \((\bar{x}, \bar{y}, \bar{z}) = (0, 0, 0)\)

Step-by-step solution

Calculate the mass of the shell

Since the density is constant, let's represent it with the symbol \(\rho\). The mass of the shell can be found by integrating the density over the shell's surface. The mass is given by: \(m = \iint_{S} \rho(x, y, z) dS\) Because the density is constant, this simplifies to: \(m = \rho \iint_{S} dS\) We can find the integral \(\iint_{S} dS\) by parameterizing the half cylinder with respect to parameters \(u\) and \(v\), such that: \(x = a\cos(u)\) \(y = v\) \(z = a\sin(u)\) where \(0 \leq u \leq \pi\) and \(-h/2 \leq v \leq h/2\). Now, we can find the partial derivatives of these parameterizations with respect to \(u\) and \(v\):

Key concepts

Surface Integrals

Surface integrals are a fundamental tool in calculus for analyzing the properties of surfaces in three dimensions. In simple terms, a surface integral can be thought of as the two-dimensional analogue of a line integral. It is a way to integrate, or sum up, quantities over a curved surface.

When calculating the surface integral of a scalar field, like mass density over a surface, the integral sums up the product of the density at each point on the surface and an infinitesimal area element at that point. The surface integral for mass is then given by the formula \( m = \iint_{S} \rho(x, y, z) dS \), where \( \rho \) is the density, and \( dS \) is the infinitesimal area element on the surface \( S \).

In the context of our exercise, the half cylinder's surface is integrated over, prescribing the scalar field of constant density to the surface, implying that every little piece of the surface contributes equally to the total mass.

Mass and Density

Mass and density are related quantities of physical significance that describe how much matter is contained in a given space. Density, denoted as \( \rho \), is a measure of mass per unit volume. In contrast, mass is the total amount of matter in an object and is a broader measure that does not consider an object's volume.

In the example with the half cylinder, the constant density means that the mass distribution is uniform across the object's surface. To find the total mass of the shell, one simple but crucial step is to compute the double integral of the density function over the surface, effectively multiplying the density by the surface area, which yields the mass: \( m = \rho \iint_{S} dS \).

Given that the density \( \rho \) is a constant, it can be factored out of the integral, simplifying the expression for mass and reducing the problem to finding the surface area \( S \) of the half cylinder.

Moment of Inertia

Moment of inertia is a physical concept related to an object's resistance to rotational acceleration about an axis. It is essentially the rotational equivalent of mass in linear motion. The larger an object's moment of inertia, the more torque is needed to attain the same rate of rotational acceleration.

Mathematically, the moment of inertia depends on the mass distribution relative to the axis of rotation. It is found by integrating the product of mass density and the square of the distance from a point on the object to the axis of rotation. In the context of our exercise, if we were to calculate the moment of inertia for the half cylinder, it would involve an integral akin to \( I = \iint_{S} \rho(x, y, z) r^2 dS \), where \( r \) is the distance from the axis.

However, in this specific exercise, we are concerned with the moments about the coordinate planes, which provide necessary components to determine the center of mass using their respective mass moments.

Parameterization

Parameterization is the process of expressing a surface as a set of equations in terms of parameters. This is very useful in multivariate calculus as it allows us to simplify calculations over complex surfaces by working with these parameters.

For our half cylinder, the parameterization means describing every point \( (x, y, z) \) on the surface in terms of two variables, \( u \) and \( v \). The parameterization given in the solution is:

• \( x = a\cos(u) \)
• \( y = v \)
• \( z = a\sin(u) \)

These equations allow us to represent all points on the half cylinder's surface using the parameters \( u \) and \( v \) within the given bounds. This approach is pivotal in turning the problem of integrating over a 3D surface into a more manageable double integral over a 2D parameter domain.

Double Integrals

Double integrals are a way to find the accumulated total of a function over a two-dimensional area. They generalize single integration, which finds the total accumulation over an interval, to two dimensions.

In practice, double integrals allow us to calculate quantities like area, volume, center of mass, and, in physics, quantities like electric charge and mass over a surface. In our textbook problem, the double integral \( \iint_{S} dS \) calculates the area of the half cylinder's surface, which is then used to find the mass of the object when multiplied by the constant density \( \rho \).

When performing a double integral via parameterization, the function we integrate (in this case, the density) is written in terms of the parameters, and the area element \( dS \) is replaced with \( |\vec{r}_u \times \vec{r}_v| dudv \) where \( \vec{r}_u \times \vec{r}_v \) is the cross product of the partial derivatives of the parameterization with respect to \( u \) and \( v \) — representing the local area scaling of the parameters.