Problem 25 - Thomas Calculus

Question

Centroid of a solid semi ellipsoid Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry three-eighths of the way from the base toward the top, show, by trans- forming the appropriate integrals, that the center of mass of a solid semiellipsoid \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)+\left(z^{2} / c^{2}\right) \leq 1, z \geq 0,\) lies on the \(z\) -axis three-eighths of the way from the base toward the top. (You can do this without evaluating any of the integrals.)

Step-by-Step Solution

Answer

The centroid is at three-eighths of the way up the z-axis, similar to the hemisphere.

Step 1: Understand the Solid Semiellipsoid

The given semiellipsoid is described by the equation \( \left(\frac{x^{2}}{a^{2}}\right) + \left(\frac{y^{2}}{b^{2}}\right) + \left(\frac{z^{2}}{c^{2}}\right) \leq 1 \) where \(z \geq 0\). It is a three-dimensional shape obtained by cutting an ellipsoid horizontally at its midpoint.

Step 2: Identify Symmetry and Volume Element

The semiellipsoid is symmetric about the \(z\)-axis. We use cylindrical coordinates to describe the volume element: \(dV = \rho \, d\rho \, d\theta \, dz\) where \(x = \rho \cos \theta\), \(y = \rho \sin \theta\), and \(z = z\). Limits for these variables are \(0 \leq \rho \leq a\), \(0 \leq \theta < 2\pi\), and \(0 \leq z \leq c\sqrt{1 - \frac{\rho^2}{a^2}}\).

Step 3: Express Volume Integral

The total volume \( V \) of the semiellipsoid is given by the integral: \[V = \int_{0}^{2\pi} \int_{0}^{a} \int_{0}^{c\sqrt{1 - \frac{\rho^2}{a^2}}} \rho \, dz \, d\rho \, d\theta\] without needing explicit evaluation, focusing on how symmetry affects the centroid.

Step 4: Integrate for Total Mass

For uniform density \( \rho_0 \), the mass \( M \) of the semiellipsoid: \[ M = \rho_0 \cdot V \] is obtained by symmetry, concentrating only on regions where integrals influence centroids distribution along the \(z\)-axis.

Step 5: Calculate the Centroid's z Coordinate

The centroid \( z_c \) is determined by the integral: \[ z_c = \frac{1}{V} \int_{0}^{2\pi} \int_{0}^{a} \int_{0}^{c\sqrt{1 - \frac{\rho^2}{a^2}}} z \rho \, dz \, d\rho \, d\theta = \frac{3}{8}c\]This shows that the centroid lies three-eighths of the distance from the base towards the top, as symmetry and earlier results for the hemisphere suggest without detailed evaluation.

Key Concepts

Solid Geometry in Semiellipsoids

Understanding solid geometry is crucial when working with three-dimensional shapes such as a semiellipsoid. A semiellipsoid is essentially half of an ellipsoid, similar to how a hemisphere is half of a sphere. In this exercise, the semiellipsoid is defined by the equation \[ \left(\frac{x^{2}}{a^{2}}\right) + \left(\frac{y^{2}}{b^{2}}\right) + \left(\frac{z^{2}}{c^{2}}\right) \leq 1 \text{ where } z \geq 0 \]This represents spatial boundaries that form the top half of an ellipsoidal shape. Recognizing these boundaries and the center of mass, or centroid, helps us understand how the shape is distributed in space.

• An ellipsoid becomes a semiellipsoid if cut horizontally at the midpoint.
• Geometry dictates the symmetry and properties of the shape.
• The centroid is a point that acts as the 'average' position of all the mass in the solid.

In practical terms, if you were to balance a semiellipsoid on a narrow rod, the centroid would be the point where it stays balanced. This exercise shows that under symmetry conditions, the centroid falls on the symmetry axis.

Symmetry in Calculus

Symmetry plays a vital role in simplifying complex calculus problems like finding the centroid of geometric shapes. In the semiellipsoid, symmetry about the \(z\)-axis means that the shape is identical on all sides as you rotate around this axis.

• Symmetry helps reduce computational workload because symmetrical properties ensure that certain integral components cancel out, simplifying calculations.
• The uniform distribution around the symmetry axis directly informs the location of the centroid.
• Calculation shortcuts emerge from recognizing these symmetrical attributes, such as deducing the z-coordinate of the centroid without explicit integral evaluation.

To exemplify symmetry, consider how a solid hemisphere's centroid is located three-eighths of the way up on the vertical axis from the base. The position of the centroid in our semiellipsoid follows a similar logic due to its symmetrical form, relying on symmetry considerations rather than extensive integral evaluation.

Introducing Cylindrical Coordinates

The use of cylindrical coordinates is an effective method for solving solid geometry problems, especially when dealing with objects like semiellipsoids. These coordinates simplify volume integration significantly when a problem involves circular symmetry.

• In cylindrical coordinates, the position of a point is defined by a radius \(\rho\), an angle \(\theta\), and a height \(z\).
• This transformation from Cartesian coordinates is particularly useful because it aligns with the inherent symmetry of the semiellipsoid.
• The boundaries in these coordinates become easier to interpret and manipulate mathematically, leading to more efficient integral set-up.

The exercise sets up the semiellipsoid's volume in cylindrical coordinates by specifying integration limits for \(\rho\), \(\theta\), and \(z\). These limits directly derive from the ellipsoid equation and the geometrical constraints, facilitating the determination of volumes and centroids without cumbersome calculations. This coordinate system exploits the rotational symmetry, making it perfect for objects like cylinders and ellipsoids, where one dimension (angle) can simply sum over its complete rotation without change.