Problem 45 - Single Variable Calculus

Question

Write an integral giving the area of the surface obtained by revolving the curve about the x-axis. $$ y=\sin x \text { on }\left[0, \frac{\pi}{2}\right] $$

Step-by-Step Solution

Answer

The integral giving the area of the surface obtained by revolving the curve \(y = \sin x\) on the interval \(\left[0, \frac{\pi}{2}\right]\) about the x-axis is: \[A = 2\pi \int_0^{\frac{\pi}{2}} \sin x \sqrt{1 + \cos^2 x} dx\]

Step 1: Calculate the derivative of the function

First, we need to find the derivative of \(y = \sin x\). To find the derivative, use the rule: \[\frac{d}{dx}(\sin x) = \cos x\] So, \(\frac{dy}{dx} = \cos x\)

Step 2: Square the derivative

Next, square the derivative: \((\frac{dy}{dx})^2 = (\cos x)^2 = \cos^2 x\)

Step 3: Input values into the surface area formula

We can now plug the values into the surface area formula: \[A = 2\pi \int_0^{\frac{\pi}{2}} \sin x \sqrt{1 + \cos^2 x} dx\]

Step 4: Evaluate the integral

Now, we will solve the integral. The integral doesn't have an elementary solution, so we will leave it in its exact form: \[A = 2\pi \int_0^{\frac{\pi}{2}} \sin x \sqrt{1 + \cos^2 x} dx\] And that's the integral giving the area of the surface obtained by revolving the curve \(y = \sin x\) on the interval \(\left[0, \frac{\pi}{2}\right]\) about the x-axis.

Key Concepts

Integral Calculus

Integral calculus is a foundational component of mathematics that involves finding and applying integrals. Integrals themselves symbolize the accumulation of quantities, such as areas under curves or the physical concept of accumulated change. This mathematics branch critically complements differential calculus, the study of rates of change and slopes of curves.

When we talk about surface area of revolution, integral calculus comes into play as we search for the accumulated surface area of a 3D object formed by rotating a 2D curve around an axis. The process consists of slicing the 3D object into infinitely thin disks or rings and then summing their areas with an integral. This is particularly complex because the area of each slice depends on its position along the curve. We harness the power of the integral to compute this continuous sum, accounting for every point on the curve to get an exact total surface area.

Surface Area Formula

The surface area formula for a surface of revolution is derived from the concept of calculating the lateral area of an infinitesimally small frustum of a cone, which approximates a small segment of the surface. This lateral area is given by 2πrh, where r is the radius from the axis of rotation to the curve and h is the slant height of the frustum segment.

When revolving a curve around the x-axis, the formula becomes \[ A = 2π \int_a^b f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx \] where f(x) represents the function of the curve, and \frac{dy}{dx} is the derivative of that function with respect to x. The expression \sqrt{1 + \left(\frac{dy}{dx}\right)^2} is the slant height segment h and 2πf(x) represents the circumference of the disk at a given point x. By integrating this expression from a to b, we evaluate the total area of the surface obtained upon revolution.

Derivative of Sine

The sine function, represented as \(y = \text{sin}(x)\), is one of the fundamental trigonometric functions in mathematics, and its derivative serves as a crucial building block in many areas of calculus and physics. The derivative of the sine function is the cosine function, symbolized mathematically as \[ \frac{d}{dx}(\sin x) = \cos x \] This relationship is particularly significant in describing wave motions and oscillations. Additionally, in the context of finding the surface area of revolution, the derivative calculates the rate at which the y-value of the sine curve changes with respect to x, which is essential for determining the slant height of the thin disk segments as the curve revolves around the x-axis. By squaring this derivative, as seen in the exercise, we make provisions for the Pythagorean identity needed to find the exact slant height in our surface area integrand.