Problem 65 - Essential Calculus

Question

Find the area of the surface generated by revolving the curve about each given axis. $$ x=4 \cos \theta, y=4 \sin \theta, \quad 0 \leq \theta \leq \frac{\pi}{2}, \quad y \text { -axis } $$

Step-by-Step Solution

Answer

The surface area of the solid generated by revolving the given curve about the y-axis is 32 square units.

Step 1: Compute the Differential of the Arc Length

First, the differential of the arc length \(ds\) needs to be computed. This can be done using the formula: \(ds=\sqrt{dx^{2}+dy^{2}}\). Because the curve is given in parametric form, \(ds\) can be written as \(\sqrt{\left(\frac{dx}{d\theta}\right)^{2}+\left(\frac{dy}{d\theta}\right)^{2}} d\theta\).

Step 2: Calculate Derivatives

To plug into \(ds\), the derivatives of \(x(\theta)\) and \(y(\theta)\) with respect to \(\theta\) are required. Given \(x = 4cos\theta\), differentiate to get \(\frac{dx}{d\theta} = -4sin\theta\). Similarly, for \(y = 4sin\theta\), the derivative is computed as \(\frac{dy}{d\theta} = 4cos\theta\). Both of these derivatives are then plugged into \(ds\).

Step 3: Simplify Integral

Upon simplification, \(ds\) equals to \(\sqrt{16}d\theta = 4d\theta\). Then, using \(x = 4cos\theta\), the surface area \(S\) is given by the integral \(S = 2 \int_a^b{x ds}\) for the y-axis of revolution. Substitute \(x\) and \(ds\) to get \(S = 2 \int_0^{\pi/2}{4 cos\theta * 4 d\theta}\).

Step 4: Solve the Integral

Computing the integral gives \(S = 32\int_0^{\pi/2}{cos\theta d\theta}\). This integrates to 32*[sin(\(\pi/2\)) - sin(0)], resulting in 32*1 = 32.

Key Concepts

Parametric Equations

Parametric equations help us to express a geometric curve by defining both x and y coordinates in terms of a third variable, often t or θ, which is known as the parameter. This method is particularly useful for describing curves that don't easily lend themselves to a simple y=f(x) or x=g(y) format.

In the context of the surface area of revolution, parametric equations allow us to trace out the curve in two dimensions as the parameter changes. For the example exercise, the curve defined by the parametric equations \(x=4\cos\theta\) and \(y=4\sin\theta\) describes a circle with radius 4, and θ varying from 0 to \(\frac{\pi}{2}\).

The use of parametric equations comes into its own when we need to compute quantities like arc length, tangents, or areas under curves, especially when revolving around an axis as is required for determining the surface area of the created shape.

Arc Length Differential

When dealing with arc lengths in calculus, the arc length differential, denoted as \(ds\), is an infinitesimal piece of the arc length that can be integrated over to find the total length of a curve between two points. In the realm of parametric equations, the differential \(ds\) is obtained from the square root of the sum of the squares of the derivatives of the parametric equations with respect to the parameter.

For our exercise, the arc length differential \(ds=\sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2}\,d\theta\) is derived from the parametric forms of x and y. Here, θ is the parameter and \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) represent the rates of change of x and y with respect to θ, respectively.

In summary, knowing the arc length differential is essential for finding the length of parametric curves and plays a crucial role in determining the surface areas of shapes created by revolving curves about an axis.

Integration in Calculus

Integration is a fundamental concept in calculus that is used to compute areas, volumes, central points, and many other useful things. It can be thought of as the reverse process of differentiation or finding the total accumulation of a quantity.

For finding the area of a surface of revolution, integration allows us to sum up infinitely many infinitely small slices of the surface to find the whole. In our given problem, the integral \(S = 2\int_0^{\pi/2}{4 \cos\theta \cdot 4 d\theta}\) involves integrating the expression that represents the area of each infinitesimally thin disc formed when the curve is revolved around the y-axis.

Why Use Integration?

Without integration, we would not have a practical method to determine such areas, especially given the complexity that arises when dealing with curves that are not straight lines. Solving the integral gives us the quantitative measure needed to describe the surface area, which is an application that showcases the power of calculus in describing and solving real-world problems.