Problem 88 - Thomas Calculus

Question

The curve \(y=\sqrt{x^{2}+1,0} \leq x \leq \sqrt{2},\) which is part of the upper branch of the hyperbola \(y^{2}-x^{2}=1,\) is revolved about the \(x\) -axis to generate a surface. Find the area of the surface.

Step-by-Step Solution

Answer

The surface area is \(\frac{2\pi (5\sqrt{2} - 4)}{3}\).

Step 1: Understand the Problem

We need to compute the surface area of the part of the hyperbola described by the function \(y = \sqrt{x^2 + 1}\) from \(x = \sqrt{1.0}\) (which simplifies to \(x = 1\)) to \(x = \sqrt{2}\) when it is revolved around the \(x\)-axis.

Step 2: Use the Surface Area Formula

To find the surface area of a curve \(y = f(x)\) when revolved about the \(x\)-axis, we use the formula:\[S = \int_a^b 2\pi y \sqrt{1 + (y')^2} \, dx\]where \(y'\) is the derivative of \(y\) with respect to \(x\).

Step 3: Find the Derivative of y

The function is \(y = \sqrt{x^2 + 1}\). The derivative of \(y\) with respect to \(x\) is:\[y' = \frac{d}{dx}\left(\sqrt{x^2 + 1}\right) = \frac{x}{\sqrt{x^2 + 1}}\]

Step 4: Set Up the Integral

Substituting \(y = \sqrt{x^2 + 1}\) and \(y' = \frac{x}{\sqrt{x^2 + 1}}\) into the surface area formula, we get:\[S = \int_{1}^{\sqrt{2}} 2\pi \sqrt{x^2+1} \sqrt{1 + \left(\frac{x}{\sqrt{x^2 + 1}}\right)^2} \, dx\]This simplifies to:\[S = \int_{1}^{\sqrt{2}} 2\pi \sqrt{x^2+1} \sqrt{1 + \frac{x^2}{x^2 + 1}} \, dx = \int_{1}^{\sqrt{2}} 2\pi \sqrt{x^2+1} \sqrt{\frac{x^2+1+ x^2}{x^2 + 1}} \, dx\]which simplifies to:\[S = \int_{1}^{\sqrt{2}} 2\pi (x^2+1) \, dx\]

Step 5: Evaluate the Integral

Solving the integral:\[S = 2\pi \int_{1}^{\sqrt{2}} (x^2+1) \, dx\]First, break it down into simpler parts:\[= 2\pi \left( \int_{1}^{\sqrt{2}} x^2 \, dx + \int_{1}^{\sqrt{2}} 1 \, dx \right)\]Calculating each integral:\[\int x^2 \, dx = \frac{x^3}{3}\quad \text{and} \quad \int 1 \, dx = x\]Evaluate these from 1 to \(\sqrt{2}\):\[\left[ \frac{x^3}{3} \right]_{1}^{\sqrt{2}} = \frac{(\sqrt{2})^3}{3} - \frac{1^3}{3} = \frac{2\sqrt{2}}{3} - \frac{1}{3}\quad \text{and} \quad [x]_{1}^{\sqrt{2}} = \sqrt{2} - 1\]Thus, the integral becomes:\[S = 2\pi \left( \frac{2\sqrt{2} - 1}{3} + \sqrt{2} - 1 \right)\]Simplifying gives:\[S = 2\pi \left( \frac{5\sqrt{2} - 4}{3} \right)\]

Step 6: Complete the Solution

The calculated surface area is:\[S = \frac{2\pi (5\sqrt{2} - 4)}{3}\]This is the surface area of the curve rotated about the \(x\)-axis.

Key Concepts

Hyperbola

In mathematics, a hyperbola is a type of conic section that can be imagined as the intersection of a double cone and a plane. A hyperbola has two branches, and in this problem, we are working with the upper branch described by the equation \( y^2 - x^2 = 1 \). This particular equation defines a standard hyperbola centered at the origin with its transverse axis along the y-axis.

For the curve we are interested in, \( y = \sqrt{x^2 + 1} \), which is part of this hyperbola, when \( y \) is solved for. The transformation into this form is based on isolating \( y \) on one side of the hyperbola equation. With these setups, the focus can then be on understanding how the properties of a hyperbola contribute to the surface geometry when rotated around an axis.

Integral Calculus

Integral calculus is a branch of mathematical analysis that deals with the accumulation of quantities and the areas under and between curves. In the context of our problem, we use it to find the surface area generated by revolving the curve around the x-axis.

The formula used to compute the surface area, \( S = \int_a^b 2\pi y \sqrt{1 + (y')^2} \, dx \), is derived from integral calculus principles. Here, we integrate across a defined interval from \( x = 1 \) to \( x = \sqrt{2} \). The integral combines both the shape specified by \( y \) and its corresponding derivative, emphasizing how integral calculus handles curve-based areas over intervals.

Derivative

The derivative is an essential concept in calculus that captures the rate of change of a function. For our problem, it's crucial because it plays a critical role in the formula used for surface area calculations.

The function \( y = \sqrt{x^2 + 1} \) requires finding its derivative concerning \( x \). Applying the chain rule, we get \( y' = \frac{x}{\sqrt{x^2 + 1}} \). This derivative reflects how the curve's slope changes and is plugged into the integral formula to adjust the evolving surface's proportional height as it's revolved around an axis.

Definite Integral

A definite integral captures the accumulation of quantities, providing a precise accumulation value over an interval. In this exercise, it helps us determine the surface area created by rotating the segment of the hyperbola from \( x = 1 \) to \( x = \sqrt{2} \) around the x-axis.

Once we set up the appropriate integral with \( S = \int_{1}^{\sqrt{2}} 2\pi \sqrt{x^2+1} \sqrt{1 + \left(\frac{x}{\sqrt{x^2 + 1}}\right)^2} \, dx \), solving it involves evaluating it over the given limits. This process breaks down into calculating the individual integrals for simplifying expressions, such as \( \int x^2 \, dx = \frac{x^3}{3} \) and \( \int 1 \, dx = x \), both evaluated from \( 1 \) to \( \sqrt{2} \). By doing so, the definite integral gives us the total surface area by summing the infinitesimally small areas generated by revolving the hyperbola segment.