Problem 79 - University Calculus: Early Transcendentals

Question

Show that the centroid of the solid semiellipsoid of revolution \(\left(r^{2} / a^{2}\right)+\left(z^{2} / h^{2}\right) \leq 1, z \geq 0\) lies on the \(z\) -axis three-eighths of the way from the base to the top. The special case \(h=a\) gives a solid hemisphere. Thus, the centroid of a solid hemisphere lies on the axis of symmetry three eighths of the way from the base to the top.

Step-by-Step Solution

Answer

The centroid of the semiellipsoid is at \(\frac{3}{8}h\) along the \(z\)-axis.

Step 1: Understand the Problem

We need to determine the centroid of a semiellipsoid described by the equation \(\left(r^2/a^2\right) + \left(z^2/h^2\right) \leq 1, z \geq 0\). The centroid should lie on the \(z\)-axis.

Step 2: Formula for Centroid

Recall that the centroid \((\bar{x}, \bar{y}, \bar{z})\) for a solid with constant density is given by \(\bar{z} = \frac{1}{V} \int z\, dV\), where \(V\) is the volume of the solid.

Step 3: Volume of the Semiellipsoid

The volume \(V\) of the semiellipsoid is \(\frac{1}{2} \times \frac{4}{3} \pi a^2 h = \frac{2}{3}\pi a^2 h \), considering just the upper half \(z \geq 0\).

Step 4: Integrating for Centroid

To find \(\bar{z}\), we use cylindrical coordinates \((r, \theta, z)\). The volume element is \(dV = r \, dr \, d\theta \, dz\). We integrate \(z\) against the volume element over appropriate limits.

Step 5: Set Limits for the Integral

The limits for \(r\) are \(0\) to \(a\sqrt{1-(z^2/h^2)}\), for \(\theta\) are \(0\) to \(2\pi\), and for \(z\) are \(0\) to \(h\).

Step 6: Solving the Integral

Solve \( \bar{z} = \frac{1}{V} \int_{0}^{2\pi} \int_{0}^{h} \int_{0}^{a\sqrt{1-(z^2/h^2)}} z r \, dr \, d\theta \, dz \). Evaluate the inner integrals for \(r\) and \(\theta\), then integrate for \(z\).

Step 7: Evaluate the Final Expression

The integral solves to \(\bar{z} = \frac{3}{8}h\), showing the centroid is at three-eighths of the height along the \(z\)-axis, as expected.

Key Concepts

Semiellipsoid

A semiellipsoid is a three-dimensional shape that resembles half of an ellipsoid. An ellipsoid is similar to a stretched sphere, and when cut in half, you get a semiellipsoid. In this exercise, we explore a semiellipsoid of revolution, meaning it is symmetric around one of its axes. This solid is defined mathematically with the equation \( \left(\frac{r^2}{a^2}\right) + \left(\frac{z^2}{h^2}\right) \leq 1 \) with \(z \geq 0\). This equation hints at its shape and the part of space it occupies. If \(h = a\), it simplifies to a hemisphere, which is a half-sphere, commonly visualized as the top half of a ball. Understanding the spatial properties of a semiellipsoid helps us grasp how it balances and where gravitational forces coincide at its centroid.

Cylindrical Coordinates

Cylindrical coordinates are a way of representing points in three-dimensional space that are especially useful when dealing with shapes like cylinders and semiellipsoids. This system uses three values:

• \(r\) - the radial distance from the z-axis,
• \(\theta\) - the angular position around the z-axis,
• \(z\) - the height above the xy-plane.

By using cylindrical coordinates, we simplify problems involving rotational symmetry, like in our semiellipsoid. It transforms the description of volume and the integration process into a more manageable form. Essentially, this approach capitalizes on the inherent symmetry to reduce complexity and focus on radial and vertical changes while revolving around a central axis.

Volume Integration

Volume integration involves calculating the total volume of a solid through integration. In finding the centroid of a semiellipsoid, this principle is crucial. Using the given semiellipsoid, we set up an integral involving the volume element \(dV = r \, dr \, d\theta \, dz\), which represents how a tiny piece of volume varies at given positions. In cylindrical coordinates, we integrate over:

• \(r\) from 0 to \(a\sqrt{1-(z^2/h^2)}\),
• \(\theta\) from 0 to \(2\pi\),
• \(z\) from 0 to \(h\).

Through this process, we find the centroid's height \(\bar{z}\) by integrating the z-value weighted by volume over the entire semiellipsoid, accounting for every infinitesimally small piece of the solid.

Symmetry Axis

The symmetry axis is a line around which a shape is invariant upon rotation, meaning the shape looks the same as it turns around this axis. In the case of our semiellipsoid, the symmetry axis is the z-axis. This specific exercise leverages the properties of symmetry to simplify calculations in determining the centroid. Because of this symmetry:

• The x and y-components of the centroid cancel out, resulting in \(\bar{x} = 0\) and \(\bar{y} = 0\).
• The centroid is constrained to lie on the z-axis.

Thanks to this geometric symmetry, finding the centroid becomes straightforward, focusing solely on the vertical distribution of volume (and thus, weight) along the z-axis. This allows us to conclude that the centroid of the semiellipsoid is at \(\frac{3}{8}h\) on the z-axis.