Problem 109 - Calculus Volume 1

Question

For the following exercises, find the volume of the solid described. The base is the region enclosed by the generic ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\). Slices perpendicular to the \(x\) -axis are semicircles.

Step-by-Step Solution

Answer

The volume is \(\frac{\pi a b^2}{6}\).

Step 1: Understand the Base of the Solid

The base of the solid is given by an ellipse described by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). This implies any point with coordinates \((x, y)\) lying on the ellipse satisfies this equation.

Step 2: Identify the Shape of the Cross-Sections

The problem states that cross-sections perpendicular to the \(x\)-axis are semicircles. Therefore, at any point \(x\) within the bounds of the ellipse, there is a semicircle above the \(x\)-axis.

Step 3: Write the Equation for the Semicircle Radius

For a given \(x\), the top and bottom points of the ellipse determine the diameter of the semicircle. The relevance is the \(y\) value from the ellipse equation: \(y = b \sqrt{1- \frac{x^2}{a^2}}\). The radius \(r\) of each semicircle is half this \(y\) value.

Step 4: Write the Area Expression for the Semicircle

The area of a semicircle is given by \(\frac{1}{2} \pi r^2\). Substitute the radius \(r = b \sqrt{1- \frac{x^2}{a^2}} / 2\) into this formula to find the area at any \(x\):\[A(x) = \frac{1}{2} \pi \left(\frac{b \sqrt{1- \frac{x^2}{a^2}}}{2}\right)^2 = \frac{\pi b^2}{8} (1-\frac{x^2}{a^2})\]

Step 5: Integrate the Area to Find the Volume

To find the volume of the solid, integrate the area of the semicircle slices along the \(x\)-axis from \(-a\) to \(a\):\[V = \int_{-a}^{a} \frac{\pi b^2}{8} \left(1-\frac{x^2}{a^2}\right) dx\]

Step 6: Solve the Integral

Evaluate the definite integral:\[V = \frac{\pi b^2}{8} \left[ \int_{-a}^{a} 1 dx - \int_{-a}^{a} \frac{x^2}{a^2} dx \right]\]Calculate each part:\[\int_{-a}^{a} 1 dx = 2a\]\[\int_{-a}^{a} \frac{x^2}{a^2} dx = \frac{2a^3}{3a^2} = \frac{2a}{3}\]Substitute back into the volume expression:\[V = \frac{\pi b^2}{8} \left(2a - \frac{2a}{3}\right) = \frac{\pi b^2 a}{6}\]

Step 7: Conclusion

Hence, the volume of the solid is \(\frac{\pi a b^2}{6}\).

Key Concepts

Ellipse

An ellipse is a fascinating geometric shape that possesses two axes of symmetry – the major axis and the minor axis. You can visualize it as a stretched or squashed circle. The standard equation of an ellipse is given by:

• \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

Here, \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes respectively. If \(a > b\), the ellipse stretches more along the x-axis, and if \(b > a\), it stretches more along the y-axis. The ellipse represents a set of all points \((x, y)\) that are equidistant from two fixed points termed as the foci. This unique property makes ellipses appear frequently in physics, astronomy, and other scientific fields.
The ellipse in this problem has its major and minor axes centered on the origin with bounds defined by the semi-major axis \(-a\) to \(a\) on x, and semi-minor axis \(-b\) to \(b\) on y. Understanding this shape is crucial as it outlines the base onto which semicircle slices are projected, forming the volume of the solid.

Semicircles

Semicircles play a pivotal role in determining the shape of the slices perpendicular to the x-axis. A semicircle is exactly half of a circle, and its boundary includes the diameter plus the half-circle line. The semicircle's equation, which governs its shape in our exercise, is linked deeply to the geometry of the ellipse.
From the ellipse equation given, \(y = b \sqrt{1 - \frac{x^2}{a^2}}\) provides the diameter of the semicircle. Since a semicircle's radius \(r\) is half of the diameter, it becomes \(\frac{b \sqrt{1 - \frac{x^2}{a^2}}}{2}\).
Keep in mind:

• The semicircles form the 'slices' or 'cross-sections' of the solid.
• Each slice's area can then be formulated as \(\frac{1}{2}\pi r^2\).

Therefore, the semicircles rising perpendicularly are governed by both the ellipse's width and the depth projection defined by the semicircle's area formula.

Calculus Integration

Integration is a fundamental concept in calculus that helps to find areas, volumes, and sums of continuous functions. In our exercise, the process of integration is used to add up an infinite number of infinitesimally small semicircle cross-sections along the x-axis to form the entire volume of the solid.
The basic idea is to set up an integral of the semicircle area expression. The expression obtained from the semicircle radius formulated to be:

• \[ A(x) = \frac{\pi b^2}{8} \left(1-\frac{x^2}{a^2}\right) \]

This equation represents the area of each semicircle slice at a certain x-position. By integrating these areas from \(-a\) to \(a\), we can calculate the full 3-dimensional volume.
Integration in this manner handles the sum of infinite small sections spread out over a specified interval, providing the accumulated total which forms the complete solid.

Definite Integral Calculation

A definite integral is a specific type of integral that calculates the net area or the accumulated quantity across a defined interval. In this exercise, we compute the exact volume of a solid via definite integration, representing both the limits and the integral.

• The required definite integral to calculate volume is:\[ \int_{-a}^{a} \frac{\pi b^2}{8} \left( 1 - \frac{x^2}{a^2} \right) dx \]

Breaking it down, we have two parts:

• The integral of the constant function 1 from \(-a\) to \(a\), \(2a\).
• The integral of \(\frac{x^2}{a^2}\) from \(-a\) to \(a\), which computes to \(\frac{2a}{3}\).

When these parts are evaluated through integration and substituted back, they provide the expected volume \(\frac{\pi a b^2}{6}\).
This definite integral consolidates and computes the slice sums into a tangible, unified volume value, allowing us to transition from mathematical abstraction to real-world application.