Problem 38 - Essential Calculus

Question

In Exercises \(35-38\), set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the \(x\) -axis. $$ y=\frac{x}{2}, \quad 0 \leq x \leq 6 $$

Step-by-Step Solution

Answer

The surface area of the solid generated by revolving the curve \( y = x/2 \) around the x-axis from \( x = 0 \) to \( x = 6 \) is approximately \( 70.68 \) square units.

Step 1: Identify the function and its derivative

The given function is \( y = x / 2 \). The derivative of the function \( y' = 1 / 2 \).

Step 2: Substitute the function and the derivative into the surface area formula

Substituting \( f(x) = x / 2 \) and \( f'(x) = 1 / 2 \) into the formula, it turns into \( A = 2 \pi \int_{0}^{6} (x/2) \sqrt{1 + (1/2)^2} dx \).

Step 3: Simplify the integrand

Inside the square root, we can simplify \( 1 + (1/2)^2 \) to \( 1.25 \), making the formula \( A = 2 \pi \int_{0}^{6} (x/2) \sqrt{1.25} dx \)..

Step 4: Pull constants out of the integral

The constants are \( \sqrt{1.25} / 2 \). Therefore the equation simplifies to \( A = 2 \pi \sqrt{1.25} / 2 \int_{0}^{6} x dx \).

Step 5: Solve the integral

Solving the integral gives \( \frac{x^2}{2}|_0^6 = 18.0 \).

Step 6: Multiply by constants outside of integral

Multiplying by outside constants gives the final surface area \( A = 2 \pi \times 18.0 \times \sqrt{1.25} / 2 = \sqrt{1.25} \pi x^2 \).

Step 7: Calculate the numerical value

Evaluating the final expression provides the numerical surface area which is approximately \( 70.68 \).

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with rates of change and accumulations of quantities. It is divided into two main parts: differential calculus and integral calculus. Differential calculus is concerned with how things change over time (derivatives), while integral calculus focuses on the accumulation of quantities (integrals).

When solving problems involving calculus, we typically find a function that represents the situation and then use calculus techniques to find derivatives or integrals, providing us with insights into the problem at hand. For example, in finding the surface area of an object obtained by revolution, we can apply integral calculus to accumulate the infinitesimal areas that compose the entire surface.

Surface Area of Revolution

The surface area of revolution is a concept in calculus that refers to the surface area of a three-dimensional object created by rotating a two-dimensional curve around an axis. The curve could be any function like the straight line in the exercise, which when rotated about the x-axis, generates a three-dimensional shape.

To calculate the surface area of such an object, we use a method that involves integral calculus. The formula involves integrating along the length of the curve, accounting for the radius from the axis of revolution and the infinitesimal distance traveled along the curve. In essence, you're summing up the areas of 'thin rings' or 'discs' that stack along the axis of rotation to form the surface.

Integration

Integration is the reverse process of differentiation, another fundamental operation in calculus. It is used to find areas, volumes, central points, and many useful things. Integral calculus gives us the tools to solve problems where we know the rate of change at each point in an interval and are looking to find the accumulated change over that interval.

By integrating the function that describes the radius of revolution as it varies along the curve's length, we obtain the surface area. For the straight-line curve in our exercise, we found that integrating the function from x = 0 to x = 6 provided the total surface area of the three-dimensional shape that resulted from the revolution. To perform this calculation, you need to understand how to evaluate the integral, which in this case is accomplished by finding the antiderivative of the function and then calculating the definite integral over the specified interval.