Problem 47 - Thomas Calculus

Question

The area of the region \(R\) enclosed by the semiellipse \(y=(b / a) \sqrt{a^{2}-x^{2}}\) and the \(x\) -axis is \((1 / 2) \pi a b\) and the volume of the ellipsoid generated by revolving \(R\) about the \(x\) -axis is \((4 / 3) \pi a b^{2} .\) Find the centroid of \(R .\) Notice that the location is independent of \(a .\)

Step-by-Step Solution

Answer

The centroid of region \( R \) is \((0, \frac{2b}{3\pi})\).

Step 1: Understanding the Problem

We need to find the centroid of the region \( R \) formed by the upper half of the ellipse \( y = \frac{b}{a} \sqrt{a^2 - x^2} \) above the \( x \)-axis. We have the area \((1 / 2) \pi a b\) and the formula for volume \((4 / 3) \pi a b^2\) when revolved around the \( x \)-axis.

Step 2: Formula for Centroid

The centroid \((\overline{x}, \overline{y})\) of a region \( R \) in the plane can be calculated using \( \overline{x} = \frac{1}{A} \int_{a}^{b} x \cdot f(x) \, dx \) and \( \overline{y} = \frac{1}{A} \int_{a}^{b} \frac{1}{2} (f(x))^2 \, dx \), where \( A \) is the area of the region \( R \).

Step 3: Find \( \overline{x} \)

For our region, \( f(x) = \frac{b}{a} \sqrt{a^2 - x^2} \). The formula for \( \overline{x} \) becomes \( \overline{x} = \frac{1}{\frac{1}{2} \pi a b} \int_{-a}^{a} x \left(\frac{b}{a} \sqrt{a^2 - x^2}\right) \, dx \). However, due to symmetry of the semiellipse about the \( y \)-axis, \( \overline{x} = 0 \).

Step 4: Simplify Finding \( \overline{y} \)

Since the area of \( R \) is \( \frac{1}{2} \pi a b \), we use \( \overline{y} = \frac{1}{A} \int_{-a}^{a} \frac{1}{2} \left(\frac{b}{a} \sqrt{a^2 - x^2}\right)^2 \, dx \). Compute this integral to get the value of \( \overline{y} \).

Step 5: Calculate \( \overline{y} \)

The integral \( \int_{-a}^{a} (\frac{b^2}{2a^2})(a^2 - x^2) \, dx \) simplifies to \( \frac{b^2}{2a^2}\left[ a^2x - \frac{x^3}{3} \right]_{-a}^{a} = \frac{b^2}{2a^2} \left( \frac{2 a^3}{3} \right) = \frac{b^2 a}{3} \). Thus, \( \overline{y} = \frac{1}{\frac{1}{2} \pi a b} \cdot \frac{b^2 a}{3} = \frac{2b}{3\pi} \).

Step 6: Conclusion

The centroid of the region \( R \) is \((0, \frac{2b}{3\pi})\). This shows that \( \overline{x} \) is zero due to symmetry and \( \overline{y} \) is independent of \( a \), satisfying the problem's condition.

Key Concepts

Semiellipse

A semiellipse is simply half of an ellipse. An ellipse itself is a smooth, symmetrical shape similar to an elongated circle. It can be described by the equation \( y = \frac{b}{a}\sqrt{a^2 - x^2} \), where \( a \) and \( b \) are the semi-major and semi-minor axes respectively. In this context, the semiellipse represents the area above the x-axis formed by that function, extending from \( -a \) to \( a \).
Understanding the semiellipse is vital for determining certain properties, like its centroid, which provides useful insights into its balance point. These properties are often useful in fields such as physics and geometry, particularly to find areas and centroids. One key multiplier of a semiellipse, especially in practical applications, is its symmetry, which allows for simplifications in calculations, notably setting its \( \overline{x} \) centroid at zero.

Revolution volumes

Revolution volumes involve creating a 3-dimensional shape by rotating a 2-dimensional area around an axis. This is an important concept in calculus used to determine the volume of complex shapes that do not have simple mathematical volume formulas.
In the context of the semiellipse problem, the semiellipse is revolved around the x-axis to create an ellipsoid. The volume of this ellipsoid can be expressed using the formula \( \frac{4}{3}\pi ab^2 \), which notably includes the radius derived from the semi-minor axis, \( b \).
The concept of revolution volumes is crucial for engineers and mathematicians who regularly deal with physical objects requiring volume calculations, such as tanks, pipes, and barrels. Understanding these volumes involves using integration to slice the shape into infinitesimally thin discs, calculating the volume of each, and then integrating across the entire shape.

Integral calculus

Integral calculus is a fundamental part of calculus focused on the process of integration, which is essentially the reverse operation of differentiation. Integration is used to find quantities like areas under curves, volumes, and centroid locations, as seen in the given problem.
Here, integral calculus aids in determining the centroid coordinates of the semiellipse. \( \overline{y} \) is found by integrating \( \frac{1}{2}((f(x))^2) \) over the specified range, using the area formula as the divisor. In the case of symmetric shapes like this semiellipse, integration simplifies because symmetry often leads to some coordinate components being zero.
Understanding integration enables solving real-world problems where determining accumulated quantities over a curve or across dimensions is necessary. By mastering integral calculus, students can calculate everything from the center of mass in physics to the total output of a function over a specific interval in mathematics.