Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the surface that results when the curve \(4 x^{2}-3 y^{2}=12\) in the \(x y\)-plane is revolved about the \(x\)-axis.

Step-by-Step Solution

Answer

The surface equation is \( 4x^2 - 3(y^2 + z^2) = 12 \).

Step 1: Identifying the Original Curve

The given curve is an equation of a hyperbola: \(4x^2 - 3y^2 = 12\). This is a hyperbola with its transverse axis along the \(x\)-axis in the \(xy\)-plane.

Step 2: Re-writing the Equation for Simplification

Rearrange the original hyperbola equation to standard form by dividing everything by 12:\[ \frac{x^2}{3} - \frac{y^2}{4} = 1. \]

Step 3: Understanding the Revolution

When the curve is revolved around the \(x\)-axis, each point on the curve traces out a circle in space with a radius equal to the \(y\)-coordinate of the original point.

Step 4: Forming the Surface Equation Upon Revolution

By revolving about the \(x\)-axis, replace \(y\) with \( \sqrt{y^2 + z^2} \) in the equation to account for the circular cross-section:\[ \frac{x^2}{3} - \frac{y^2 + z^2}{4} = 1. \]

Step 5: Final Equation of the Surface

The surface formed is an equation: \[ 4x^2 - 3(y^2 + z^2) = 12. \] This is the required equation representing the surface formed when the hyperbola is revolved around the \(x\)-axis.

Key Concepts

Equation of a surface

In calculus, the equation of a surface is a mathematical expression that describes all the points lying on a geometric surface in three-dimensional space. A surface can be seen as an extension of two-dimensional curves into an extra dimension.
For this exercise, we start with a hyperbola, which is a type of conic section represented by the equation: \(4x^2 - 3y^2 = 12\).
We are tasked with finding the three-dimensional surface resulting from revolving this hyperbola around the \(x\)-axis.
To accomplish this, you convert the two-dimensional equation of the curve into a three-dimensional surface equation.

• First, it is useful to restate the hyperbola in its standard form to make subsequent steps more straightforward: \(\frac{x^2}{3} - \frac{y^2}{4} = 1\).
• Next, since we are revolving the curve about the \(x\)-axis, we incorporate an additional variable, \(z\), to transform the original curve into a surface.

Hyperbola

A hyperbola is a type of conic section formed by the intersection of a plane with both nappes of a double cone. It appears as two separate, symmetrical curves.
Every hyperbola has certain characteristics:

• Two branches that open either vertically or horizontally depending on its orientation.
• A center, which is the point equidistant from foci of the hyperbola.
• Asymptotes, which are straight lines that the branches approach but never touch.
• A transverse axis, the line segment joining the vertices.

In our exercise, the original equation \(4x^2 - 3y^2 = 12\) represents a hyperbola in the \(xy\)-plane with its transverse axis along the \(x\)-axis.
Rearranging it to the standard form: \(\frac{x^2}{3} - \frac{y^2}{4} = 1\), we can more clearly identify its properties. This hyperbola is centered at the origin and its transverse axis is parallel to the \(x\)-axis.

Revolution around an axis

When a curve is revolved around an axis, it creates a surface in three-dimensional space. This process involves rotating every point on the curve by 360 degrees around the axis.
The original problem requires us to revolve a hyperbola about the \(x\)-axis.

• The revolution converts each point \((x, y)\) on the hyperbola into a circle of radius equal to \(y\), lying in a plane perpendicular to the axis of revolution.
• In mathematical terms, this is achieved by replacing the \(y\) coordinate in the hyperbola equation with \(\sqrt{y^2 + z^2}\), which accounts for all the points in the circular cross-section.

Carrying out this transformation gives us the surface equation: \(4x^2 - 3(y^2 + z^2) = 12\).
This equation describes a hyperboloid of one sheet, which is the three-dimensional surface obtained after revolving the hyperbola around the \(x\)-axis.
This elegant surface depicts a symmetric shape extending in three dimensions, perfectly illustrating how calculus can transform simple curves into complex geometrical structures.