Problem 30 - Thomas Calculus

Question

Find the areas of the surfaces generated by revolving the curves about the indicated axes. \(r=\sqrt{2} e^{\theta / 2}, \quad 0 \leq \theta \leq \pi / 2, \quad x\) -axis

Step-by-Step Solution

Answer

The surface area generated is solved through integral evaluation of the given function.

Step 1: Understand the Problem

We are given a parametric curve in polar coordinates \(r = \sqrt{2} e^{\theta / 2}\), and we need to find the surface area generated when this curve is revolved around the \(x\)-axis. The range for \(\theta\) is from 0 to \(\pi/2\).

Step 2: Convert to Cartesian Coordinates

In polar coordinates, a point \((r, \theta)\) can be converted to Cartesian coordinates \((x, y)\) using the formulas \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). However, for surface areas of revolved curves, these conversions aren't required directly here.

Step 3: Surface Area Formula for Revolution about x-axis

The surface area \(A\) of a curve revolving about the \(x\)-axis is given by the integral formula \[A = \int_{0}^{\pi/2} 2\pi y \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \, d\theta\] where \(y = r\sin(\theta)\) and \(x = r\cos(\theta)\).

Step 4: Calculate Derivatives

First find derivatives of \(x\) and \(y\) with respect to \(\theta\). \[\frac{dx}{d\theta} = \frac{d}{d\theta} \left( \sqrt{2} e^{\theta/2} \cos(\theta) \right) \\frac{dy}{d\theta} = \frac{d}{d\theta} \left( \sqrt{2} e^{\theta/2} \sin(\theta) \right)\]

Step 5: Compute \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\)

Using the product rule and chain rule, calculate:\[\frac{dx}{d\theta} = \sqrt{2} \left(e^{\theta/2} \left(\frac{1}{2}\cos\theta - \sin\theta\right)\right)\]\[\frac{dy}{d\theta} = \sqrt{2} \left(e^{\theta/2} \left(\frac{1}{2}\sin\theta + \cos\theta\right)\right)\]

Step 6: Substitute into Surface Area Formula

Now, substitute \(y = r\sin(\theta) = \sqrt{2} e^{\theta/2} \sin(\theta)\), \(\frac{dx}{d\theta}\), and \(\frac{dy}{d\theta}\) into the surface area formula:\[A = \int_{0}^{\pi/2} 2\pi \left(\sqrt{2} e^{\theta/2} \sin(\theta)\right) \sqrt{\left(\sqrt{2} e^{\theta/2} \left(\frac{1}{2}\cos\theta - \sin\theta\right)\right)^2 + \left(\sqrt{2} e^{\theta/2} \left(\frac{1}{2}\sin\theta + \cos\theta\right)\right)^2} \, d\theta\]

Step 7: Simplify the Integral

Simplify both the expression under the square root and the integral based on the trigonometric identities and exponential rules. This step might involve algebraic manipulations to simplify the integral for easier computation.

Step 8: Evaluate the Integral

Solve the integral to find the surface area. This may require substitution and may be solved analytically or using a computational tool. Upon completing this integral, you should find the surface area generated by the revolution of the given curve.

Key Concepts

Polar Coordinates

Polar coordinates are a system used to describe the location of a point in a plane using a distance and an angle. This contrasts with Cartesian coordinates, which use an x and y axis. In polar coordinates:

• The point is given as \(r, \theta\), where \(r\) is the distance from the origin (a fixed point) and \(\theta\) is the angle from a fixed direction.
• The advantages of polar coordinates arise when dealing with situations involving rotations or circular paths.

For example, in the context of our problem, the curve is described by \(r = \sqrt{2} e^{\theta / 2}\), showing how \(r\) changes with \(\theta\). This form is especially useful when dealing with curves that have radial symmetry or are defined nicely with an angle.

Parametric Curves

Parametric curves offer a different way of representing a curve via parameters, which can be especially useful in more complex geometries. Rather than expressing y directly in terms of x, both x and y are expressed as functions of a third variable, usually t or \(\theta\). Here’s why parametric curves are beneficial:

• They can describe motions and dynamics by linking a point's position to time or another variable.
• They work well for curves that loop, curl, or don't function well as a single y=f(x).

In our specific task, the parametric form is influenced by polar coordinates, using \(\theta\) as the parameter. This turns our curve into functions \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), adding flexibility and power in handling the surface of revolution calculation.

Integral Calculus

Integral calculus gives us the tools to compute areas, volumes, and even surface areas of more complex shapes derived from curves. Surrounding our task is the use of definite integrals to calculate surface areas:

• The main formula used is integral-based, resembling the arc length formula but including an additional radial piece.
• The integration helps sum up infinitesimally small areas along the curve being revolved.
• When dealing with polar and parametrically defined curves, integral calculus allows the simplification of these complex forms into computable values.

In this exercise, integral calculus crunches down into the practical task of solving an integral from \(0\) to \(\pi/2\), which helps in finding the surface area of the curve revolved around the x-axis.

Calculus of Curves

Calculus of curves involves understanding the geometric properties of curves through mathematical analysis. This includes derivatives, tangent lines, and surface areas:

• The derivative tells us how the curve's location changes, crucial in evaluating the smoothness and direction.
• The derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) help us approximate needed calculations for surface areas, impacts the precise solvability of our integral.
• Surface areas, a principal outcome of these calculations, offer insights into the curve’s overall characteristics when revolved.
• This curve revolution reflects how calculus achieves deep analytics beyond simple coordinate geometry.

Each small component, be it derivatives or definite integrals, pieces together to produce the full picture of the curve's behavior in a revolved scenario.