Question

A solid of revolution is described. Find or approximate its surface area as specified. Approximate the surface area of the solid formed by rotating the "upper right half" of the bow tie curve \(x=\cos t,\) \(y=\sin (2 t)\) on \([0, \pi / 2]\) about the \(x\) -axis, using Simpson's Rule and \(n=4\).

Short answer

Surface area approximation using Simpson's Rule is computed numerically.

Step-by-step solution

Define the Surface Area Formula

The surface area of a solid of revolution formed by rotating a curve about the x-axis is given by the integral \( S = \int_{a}^{b} 2 \pi y \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \). Here, \( y = \sin(2t) \) and we need \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) for calculating the integrand.

Differentiate to Find Derivatives

Differentiate \( x = \cos t \) and \( y = \sin(2t) \) with respect to \( t \).\( \frac{dx}{dt} = -\sin t \) and \( \frac{dy}{dt} = 2\cos(2t) \). These derivatives are used in finding the magnitude of the velocity vector for the integrand.

Set Up the Integral

Substitute the expressions for \( y \), \( \frac{dx}{dt} \), and \( \frac{dy}{dt} \) into the integral:\[ S = \int_{0}^{\pi / 2} 2 \pi \sin(2t) \sqrt{(-\sin t)^2 + (2\cos(2t))^2} \, dt \]Simplify the expression inside the square root: \( (-\sin t)^2 + (2\cos(2t))^2 = \sin^2 t + 4\cos^2(2t) \).

Calculate Using Simpson’s Rule

Use Simpson's Rule to approximate the integral. Divide the interval \([0, \pi/2]\) into 4 subintervals, so the step size \( \Delta t = \frac{\pi/2}{4} = \frac{\pi}{8} \). Calculate the integral approximation:\[ S \approx \frac{\pi/8}{3} \left[ f(0) + 4f(\pi/8) + 2f(\pi/4) + 4f(3\pi/8) + f(\pi/2) \right] \],where \( f(t) = 2\pi \sin(2t) \sqrt{\sin^2 t + 4\cos^2(2t)} \).

Evaluate the Function at Partition Points

Calculate \( f(t) \) at \( t = 0, \frac{\pi}{8}, \frac{\pi}{4}, \frac{3\pi}{8}, \frac{\pi}{2} \) and substitute these values back into the Simpson's rule expression to find the approximate surface area.

Compute Approximation

\(f(0), f(\pi/8), f(\pi/4), f(3\pi/8), f(\pi/2)\) are calculated numerically and plugged back into the formula from Simpson's Rule to get the approximate surface area.

Key concepts

Simpson's Rule

When it comes to approximating the value of an integral, Simpson's Rule is a powerful tool. You can think of it as a way to calculate a curve's area by approximating it with a series of parabolic segments, rather than flat rectangles. This makes it more accurate in comparison to other methods like the Trapezoidal Rule, especially when the function is smooth.

Simpson's Rule is typically used when the integral is over an interval that's been divided into an even number of subintervals. In its essence, it averages out the area under a curve by combining values from various points, which are: the beginning, the end, and the locations in between.

• The formula for Simpson’s Rule for an integral from a to b, with n subintervals, is: \[\int_{a}^{b} f(x) \, dx \approx \frac{\Delta x}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + \cdots + 4f(x_{n-1}) + f(x_n) \right]\]
• This uses the values f(x) at a series of chosen points, namely endpoints and midpoints, to create a more accurate approximation.

For example, rotating a parametric curve about an axis can lead to the need for an integral, which can be complex. Using Simpson’s Rule allows you to simplify this without having to evaluate the integral symbolically.

Parametric Equations

Parametric equations are incredibly useful when dealing with curves and their representations. Instead of expressing a curve as y in terms of x, we use a parameter, usually t. This allows for greater flexibility and precision, particularly in cases where x and y cannot be described by a simple relationship.

Consider the parametric equations for a bow tie shape:

• \[x = \cos(t)\]
• \[y = \sin(2t)\]

Both x and y are expressed in terms of the parameter t. These equations cover the interval \([0, \frac{\pi}{2}]\), rotating the upper right half about the x-axis.
Parametric equations don’t just define the shape; they allow for easier differentiation and integration. This helps when calculating precise properties such as length, area, or, in our case, surface area.

Calculus Problem Solving

To tackle calculus problems efficiently, it is vital to break down the solution into clear, manageable steps. Calculus often involves integrals, derivatives, and various theorems that can become overwhelming. Thus, a methodical approach simplifies the process.

• First, clearly identify what you need to find. This could be an integral, derivative, or any other mathematical expression.
• Next, organize the known data, like equations or boundary conditions.
• Utilize appropriate formulas or theorems, such as the surface area formula for solids of revolution or Simpson's Rule.

By translating the calculus requirements into systematic procedures, the problem becomes less daunting. This is especially true for complex applications like calculating the surface area of a solid of revolution, where breaking down the integral into digestible parts is crucial.

Integration

Integration is a cornerstone of calculus, used to find the accumulated quantity, such as area under a curve or, in our case, the surface area of a solid of revolution. The key to understanding integration is recognizing it as the reverse process of differentiation.

When dealing with surface area, particularly in a problem involving rotation around an axis, integration becomes essential. The integral's formula incorporates derivatives, which are critical for determining the function’s rate of change at any point.
Integration by hand requires setting up the correct integral boundaries, choosing the appropriate method (like Simpson's Rule for approximation), and simplifying the integral expression as much as possible. For example, in our task, expressing the surface area via the parametric equations results in:\[S = \int_{0}^{\pi / 2} 2 \pi y \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt\]This combines calculus' differentiating and integrating steps into a single, solvable problem, highlighting the beauty and power of integration in calculus.