Problem 101 - Thomas Calculus

Question

Centroid of solid semiellipsoid 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 a solid semiellipsoid lies at \((0, 0, \frac{3h}{8})\).

Step 1: Understand the Problem

Recognize that we are tasked with finding the centroid of a solid semiellipsoid formed by revolving an ellipse with the equation \(\frac{r^2}{a^2} + \frac{z^2}{h^2} \leq 1, \) where \(z \geq 0 \), about the \(z\)-axis. This centroid should be found at a point \( \left(0, 0, \frac{3h}{8} \right) \) along the \(z\)-axis.

Step 2: Set Up the Volume Element

The volume element \(dV\) of the semiellipsoid using cylindrical coordinates is given by \( dV = r \, dr \, d\theta \, dz \). We will integrate \(dV\) to find the volume \(V\) of the semiellipsoid.

Step 3: Integrate to Find the Volume

Express the limits for \(r\), \(\theta\), and \(z\): \(0 \leq r \leq a \sqrt{1 - \frac{z^2}{h^2}}\), \(0 \leq \theta < 2\pi\), and \(0 \leq z \leq h\). Integrate \(dV\) over these boundaries to calculate the total volume \(V\). \[V = \int_{0}^{h} \int_{0}^{2\pi} \int_{0}^{a \sqrt{1 - \frac{z^2}{h^2}}} r \, dr \, d\theta \, dz\]

Step 4: Evaluate the Volume Integral

First, integrate with respect to \(r\): \[\int_{0}^{a\sqrt{1-\frac{z^2}{h^2}}} r \, dr = \frac{1}{2} \left[a^2(1-\frac{z^2}{h^2})\right]\]Then integrate with respect to \(\theta\):\[\int_{0}^{2\pi} \, d\theta = 2\pi\]Finally, integrate with respect to \(z\): \[V = \pi a^2 \int_{0}^{h} (1-\frac{z^2}{h^2}) \, dz\]

Step 5: Determine the Center of Mass

To find the \(z\)-coordinate of the centroid \(\bar{z}\), express it as:\[\bar{z} = \frac{1}{V} \int_{0}^{h} \int_{0}^{2\pi} \int_{0}^{a \sqrt{1-\frac{z^2}{h^2}}} z r \, dr \, d\theta \, dz\]Carry out similar integrals as before to evaluate this expression.

Step 6: Evaluate the Integrals for \(\bar{z}\)

First integrate with respect to \(r\):\[\int_{0}^{a\sqrt{1-\frac{z^2}{h^2}}} z r \, dr = \frac{a^2z}{2} \left(1-\frac{z^2}{h^2}\right)\]Then integrate with respect to \(\theta\):\[\int_{0}^{2\pi} \, d\theta = 2\pi\]Finally, integrate with respect to \(z\): \[\bar{z} = \frac{\pi a^2}{V} \int_{0}^{h} z(1-\frac{z^2}{h^2}) \, dz\]Completing these integrals will show that \(\bar{z} = \frac{3h}{8}\).

Step 7: Verification for Hemisphere

When \(h = a\), the semiellipsoid becomes a hemisphere. Evaluate the integrals with \(h = a\) to verify that the centroid lies at \(\bar{z} = \frac{3a}{8}\), consistent with the result for the semiellipsoid.

Key Concepts

Centroid calculation

When dealing with geometric objects in calculus, one often needs to determine the centroid, or the geometric center, of a solid. The centroid can be thought of as the point that represents the center of mass for a uniform density object. In this exercise, we seek the centroid of a solid semiellipsoid, a three-dimensional shape resulting from the rotation of an ellipse around one of its axes. The calculation involves averaging the coordinates of all the infinitesimal volume elements that make up the shape.

For our semiellipsoid, the centroid is expected to lie on the axis of symmetry, the z-axis, due to symmetry considerations. To find it, we use integration to sum up the positions of all these elements and divide by the total volume of the semiellipsoid. The crucial step is solving the integral expression for the z-coordinate of the centroid, which involves integrating position functions with respect to volume elements. Once complete, our calculations show that the centroid is \(\bar{z} = \frac{3h}{8}\), confirming that it is three-eighths of the way from the base to the top.

Cylindrical coordinates integration

To solve problems involving solids of revolution, like the semiellipsoid, utilizing cylindrical coordinates is highly advantageous. These coordinates (r, \(\theta\), z) are particularly useful for objects symmetric around an axis, such as our semiellipsoid around the z-axis.

Cylindrical coordinates transform the volume element into \(dV = r \, dr \, d\theta \, dz\), making it simpler to express and solve integrals in three-dimensional spaces that possess rotational symmetry. The limits of integration for r, \(\theta\), and z are carefully determined based on the shape of the semiellipsoid.

• \(r\) ranges between 0 and \(a \sqrt{1 - \frac{z^2}{h^2}}\)
• \(\theta\) goes from 0 to \(2\pi\)
• z varies from 0 to h

By integrating the volume element over these limits, we can calculate both the total volume of the solid and the coordinates of its centroid. This method simplifies the ordinarily complex task of volume and mass distribution calculations in non-standard shapes.

Volume of semiellipsoid

Finding the volume of a semiellipsoid requires integrating the volume element over the entire region occupied by the solid. In cylindrical coordinates, this process involves integrating the function \(r\) over the specified range of r, \(\theta\), and z.

The integration is broken down as follows:

• First, integrate with respect to \(r\), which accumulates the volume across concentric cylindrical shells.
• Next, integrate with respect to \(\theta\), effectively wrapping the shells around the axis.
• Finally, integrate with respect to \(z\), summing these shells along the height of the semiellipsoid.

By progressing systematically through each coordinate, we compute the total volume, \(V\), accurately. This calculated volume becomes a fundamental part of finding the centroid, as it normalizes the summed positions of the volume elements to derive the centroid location. The formulas result in a volume consistent with known outcomes for semiellipsoids, linking geometric theory with practical calculus.

Center of mass

Finding the center of mass involves calculating where a hypothetical point could balance the solid, assuming uniform density. For a solid like our semiellipsoid, this point is crucially the same as the centroid due to constant density.

The z-coordinate of this center, \(\bar{z}\), is determined using integration. The integral of \(z\) across the semiellipsoid, adjusted for density, provides this value. Again, cylindrical coordinates are employed to streamline calculations, simplifying the representation of the semiellipsoid.

• Calculate the integral of the weighted volume element \(z \, r \, dr \, d\theta \, dz\).
• Divide the result by the total volume \(V\) to find \(\bar{z}\).

The resultant expression proves \(\bar{z} = \frac{3h}{8}\), indicating the center of mass lies three-eighths of the way along the z-axis from the base of the semiellipsoid. This result also confirms symmetry considerations, as the mass is evenly distributed on either side.

Solid geometry in calculus

In calculus, solid geometry involves calculating properties of three-dimensional shapes using integration methods, a fundamental concept for understanding physical and engineering principles. Calculus provides tools to determine not just centroids but volumes, surface areas, and other critical properties of geometric solids, such as the semiellipsoid.

When applied to geometrically complex shapes, calculus allows us to find elusive characteristics by representing the shapes as integrals. For solids of revolution, these integrals often involve re-aligning our theoretical point of view through transformed coordinate systems, like cylindrical or spherical coordinates.

By rigorously applying these methods, we connect geometric understanding with analytical rigor, leading to derivations and formulas that define core shape properties. Practically speaking, these calculations bridge abstract mathematical concepts with real-world application, critical in physics and beyond, because these methods provide a linear progression from simple shapes to complex real-life models.