Problem 4 - Thomas Calculus

Question

In Exercises \(1-8 :\) a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. b. Graph the curve to see what it looks like. If you can, graph the surface, too. c. Use your grapher's or computer's integral evaluator to find the surface's area numerically. \(x=\sin y, \quad 0 \leq y \leq \pi ; \quad y\) -axis

Step-by-Step Solution

Answer

Revolve \( x = \sin y \) about \( y \)-axis to find surface. Integral \( \int_{0}^{\pi} 2\pi \sin y \sqrt{1+\cos^2 y} \, dy \). Use tool to get numeric area.

Step 1: Understand the Problem

We need to find the surface area of the surface generated by revolving the curve \( x = \sin y \) about the \( y \)-axis, given that \( 0 \leq y \leq \pi \).

Step 2: Set Up the Integral for Surface Area

The formula for the surface area of a surface of revolution about the \( y \)-axis is given by \( A = \int_{a}^{b} 2\pi x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy \). Here, \( x = \sin y \), so \( \frac{dx}{dy} = \cos y \). The bounds are \( 0 \) to \( \pi \) for \( y \). Thus, the integral becomes \( A = \int_{0}^{\pi} 2\pi \sin y \sqrt{1 + \cos^2 y} \, dy \).

Step 3: Graph the Curve

Graph the curve \( x = \sin y \) for \( y \) ranging from 0 to \( \pi \). The curve starts at point (0, 0) and arcs upwards to (0, \( \pi \)) coinciding with well-known sine wave segments.

Step 4: Evaluate the Integral Numerically

Using a graphing calculator or computer software, evaluate the integral \( \int_{0}^{\pi} 2\pi \sin y \sqrt{1 + \cos^2 y} \, dy \). For example, using a tool like Wolfram Alpha or a calculator, find the approximate numeric value of the surface area.

Key Concepts

Integral Setup

When faced with finding the surface area of a curve revolved around an axis, setting up the integral is key. In this exercise, the curve given is \( x = \sin y \) and it is revolved around the \( y \)-axis over the interval \( 0 \leq y \leq \pi \).

The formula used for the surface area \( A \) when revolving about the \( y \)-axis is: \[ A = \int_{a}^{b} 2\pi x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \ dy \] Here:

• \( 2\pi x \) is the circumference of the infinitesimal circle made when \( x \) at a particular \( y \) is revolved around the \( y \)-axis.
• \( \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \) arises from the arc length formula, accounting for the contour of the curve.
• The limits of integration are \( a = 0 \) and \( b = \pi \), matching the range of \( y \).
• For \( x = \sin y \), \( \frac{dx}{dy} = \cos y \).

Therefore, the integral is: \[ A = \int_{0}^{\pi} 2\pi \sin y \sqrt{1 + \cos^2 y} \ dy \] This integral neatly sets the stage for numerical evaluation and further exploration.

Numerical Integration

Now that the integral for the surface area is set up, we move to evaluating it numerically. Often, complex integrals do not have simple antiderivatives, making numerical methods essential. Tools such as graphing calculators, Excel, or software like WolframAlpha or MATLAB can be used to approximate these integrals.

For our integral: \[ \int_{0}^{\pi} 2\pi \sin y \sqrt{1 + \cos^2 y} \ dy \] A numerical integrator will approximate the value of this integral over the interval by subdividing the region into smaller segments and calculating the area of these segments. They do this by methods such as:

• Trapezoidal Rule - Approximates the region under the curve as a series of trapezoids.
• Simpson's Rule - Uses parabolic arcs instead of straight lines for more accuracy.

Using these methods, you can get an accurate numeric solution to the integral, providing the desired surface area. This process helps in visualizing how different parts of the curve contribute to the total area.

Graphing Technique

Visualizing the problem through graphing is an invaluable technique. Not only does it clarify the shape and region involved, but it also aids in verifying your setup and understanding the problem more holistically.

Here's how to approach graphing for this exercise:

• Graph \( x = \sin y \) over the interval \( 0 \leq y \leq \pi \). The graph will start at \((0,0)\) and rise to \((0,\pi)\), forming part of the well-known \(\sin\) curve.
• Consider how this curve looks when revolved around the \( y \)-axis.
• Graphing software can help visualize this surface of revolution by rotating the \(\sin y\) along the \( y\)-axis. Often, in 3D graphers, we can visually simulate the revolution.

Seeing the graph helps to:

• Conceptualize the integral setup by knowing the bounds and revolution path.
• Detect any errors in understanding such as missed features.
• Appreciate the symmetry and shape that generate from the revolution, aiding in understanding the geometry of the problem.

Graphing isn't just about drawing; it's about connecting the analytical setup with a visual understanding.