Question

Let \(D\) be the region bounded by the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1,\) where \(a > 0, b > 0\) and \(c>0\) are real numbers. Let \(T\) be the transformation \(x=a u, y=b v, z=c w\). Find the center of mass of the upper half of \(D(z \geq 0)\) assuming it has a constant density.

Short answer

Answer: The coordinates of the center of mass of the upper half of the ellipsoid are given by \((x_{cm}, y_{cm}, z_{cm})\), which can be found using the following expressions: $$ (x_{cm}, y_{cm}, z_{cm}) = \left(\frac{1}{V}\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} a^2u\, dw\, dv\, du, \frac{1}{V}\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} b^2v\, dw\, dv\, du, \frac{1}{V}\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} c^2w\, dw\, dv\, du \right) $$ where V is the volume of the region.

Step-by-step solution

Define the upper half of the ellipsoid in terms of u, v, and w

We are given the transformation \(x = au, y = bv, z = cw\). Substitute the values of x, y, and z in the original equation. \((au)^2/a^2 + (bv)^2/b^2 + (cw)^2/c^2 = 1\) Solve for z in terms of u, v, and w: \(w^2 = 1 - u^2 - v^2\) with \(w \geq 0\)

Calculate the volume and mass of the region

The region is given in terms of u, v, and w. To find the volume, use a triple integral in these coordinates. The Jacobian of transformation is the determinant of the matrix formed by the partial derivatives from the transformation. $$ \text{Jacobian} = \left|\begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{matrix}\right| = \left|\begin{matrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{matrix}\right| = abc $$ The limits of integration for u and v are -1 to 1, and for w, since it is non-negative, from 0 to \(\sqrt{1 - u^2 - v^2}\). The volume V is given by: $$ V = \int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} abc\, dw\, dv\, du $$ Assuming a constant density, the mass M is equal to the density times the volume.

Find the coordinates of the center of mass

The coordinates (x, y, z) of the center of mass are given by: $$ x_{cm} = \frac{\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} ax (abc)\, dw\, dv\, du}{abcV} = \frac{1}{V}\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} a^2u\, dw\, dv\, du $$ $$ y_{cm} = \frac{\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} by (abc)\, dw\, dv\, du}{abcV} = \frac{1}{V}\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} b^2v\, dw\, dv\, du $$ $$ z_{cm} = \frac{\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} cz (abc)\, dw\, dv\, du}{abcV} = \frac{1}{V}\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} c^2w\, dw\, dv\, du $$

Write the final result

The center of mass of the upper half of the ellipsoid is given by the coordinates \((x_{cm}, y_{cm}, z_{cm})\). From the calculations above, we find: $$ (x_{cm}, y_{cm}, z_{cm}) = \left(\frac{1}{V}\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} a^2u\, dw\, dv\, du, \frac{1}{V}\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} b^2v\, dw\, dv\, du, \frac{1}{V}\int_{-1}^1 \int_{-1}^1 \int_{0}^{\sqrt{1-u^2-v^2}} c^2w\, dw\, dv\, du \right) $$

Key concepts

Triple Integral

A triple integral is like a treasure map for finding the volume under consideration. It's used to compute volumes of three-dimensional objects by integrating over three variables. When you perform a triple integral, you typically integrate in three dimensions using variables like \(x\), \(y\), and \(z\), or in our transformed space, \(u\), \(v\), and \(w\).
For example, to find the volume of the upper half of our ellipsoid, we apply a triple integral over these three variables. The limits of integration are derived from the space in which the ellipsoid resides:
- The limits for \(u\) and \(v\) range from -1 to 1 because they represent scaled versions of \(x\) and \(y\) with radii normalized to 1.- The limit for \(w\) ranges from 0 to \(\sqrt{1 - u^2 - v^2}\) because this represents the upper hemisphere (\(z \geq 0\)).
This integral effectively "adds up" the density over all the tiny volumes within the ellipsoid to find the total volume or mass, assuming constant density.

Jacobian Transformation

The Jacobian transformation is a crucial step when changing variables in integrals. It deals with how small changes in the transformed variables affect the original variables. Here, it helps convert an integral in \((x, y, z)\) space to an integral in \((u, v, w)\) space.
In our problem, you will notice that we need to compute a Jacobian determinant when transforming from \( (x, y, z) \) to \( (u, v, w) \). This determinant is:
\[\text{Jacobian} = \left|\begin{matrix}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w}\end{matrix}\right| = abc\]
This tells us how the scales of \(u\), \(v\), and \(w\) are stretched to fill the original \(x\), \(y\), and \(z\) space. The Jacobian determinant \(abc\) is then used as a multiplication factor in the triple integral, ensuring we correctly calculate the volume based on the transformed variables.

Ellipsoid

An ellipsoid is like a stretched sphere. It is a three-dimensional shape represented mathematically by the equation:
\[\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\]
The numbers \(a, b,\) and \(c\) describe how much the ellipsoid is stretched along the \(x, y\) and \(z\) axes, respectively. Think of them as radii along each axis:
- \(a\) is the stretching radius along the \(x\)-axis.- \(b\) is the stretching radius along the \(y\)-axis.- \(c\) is the stretching radius along the \(z\)-axis.
In the problem, we specifically focus on the upper half of this ellipsoid, where \(z \geq 0\). This means we only want the volume above the \(z = 0\) plane, which gives us a hemisphere. By considering the transformation \(T\) given by \(x = au, y = bv, z = cw\), we normalize the ellipsoid to a unit sphere in the \(u, v, w\) space, simplifying our work.

Constant Density

Density is a measure of how much mass is present in a given volume. When we say an object has a constant density, it implies that the mass is evenly distributed across the entire volume. In simpler terms, every small portion of the object has exactly the same amount of mass.
For the upper half of the ellipsoid in our problem, assuming constant density means that we can multiply this density by the calculated volume to determine the total mass. The beauty of this assumption is that it simplifies calculations related to the center of mass, because the mass distribution does not vary with location within the object.
If the density were not constant, the location of the center of mass could shift, adding additional complexity to both the problem setup and mathematical treatment. In contrast, with constant density, the center of mass depends solely on the geometry and bounds of the volume, not on varying mass concentrations.