Question

Let \(f\) be differentiable and positive on the interval \([a, b] .\) Let \(S\) be the surface generated when the graph of \(f\) on \([a, b]\) is revolved about the \(x\) -axis. Use Theorem 14.12 to show that the area of \(S\) (as given in Section 6.6 ) is $$ \int_{a}^{b} 2 \pi f(x) \sqrt{1+f^{\prime}(x)^{2}} d x $$.

Short answer

Question: Prove that the formula for the area of a surface generated by revolving a curve around the x-axis agrees with the given formula: $$S = \int_a^b 2 \pi f(x) \sqrt{1+f'(x)^2} dx$$ Answer: To prove the given formula, we followed these steps: 1. Reviewed Theorem 14.12 that provides the formula for the length of a curve given by a vector function. 2. Reviewed the general area formula for the surface of revolution: $$S = \int_a^b 2 \pi f(x) ds$$ 3. Determined the arc length (ds) in terms of x and f'(x) using the parameterization r(t) = and Theorem 14.12, which resulted in: $$ds = \sqrt{1+f'(t)^2} dt$$ 4. Substituted the found ds into the surface area formula and obtained the given formula: $$S = \int_a^b 2 \pi f(x) \sqrt{1+f'(x)^2} dx$$ Thus, we proved that the area formula for the surface generated by revolving a curve around the x-axis agrees with the provided formula.

Step-by-step solution

Review Theorem 14.12

Theorem 14.12 states that if a curve C is given by the vector function r(t) = , a ≤ t ≤ b, where x(t), y(t), and z(t) have continuous derivatives on [a, b], then the length of C is given by $$L=\int_{a}^{b} \sqrt{x'(t)^{2}+y'(t)^{2}+z'(t)^{2}} dt$$.

Review the area formula for a surface of revolution

The formula for the area of a surface generated when a graph on the interval \([a,b]\) is revolved about the \(x\)-axis is: $$S = \int_a^b 2 \pi f(x) ds$$, where ds is the arc length of the curve.

Find the arc length of the curve in terms of x and f'(x)

We need to determine the length of the curve in terms of x and f'(x). To do this, let's consider the parameterization r(t) = . Then x = t, y = f(t), and z = 0. Now we calculate the derivatives with respect to t: x'(t) = 1 \\ y'(t) = f'(t) \\ z'(t) = 0 Using Theorem 14.12, we find the length: $$ds = \sqrt{1^2 + f'(t)^2 + 0^2} dt = \sqrt{1+f'(t)^2} dt$$

Plug ds into the surface area formula

Now that we have ds in terms of x and f'(x), we can substitute it into the surface area formula: $$S = \int_a^b 2 \pi f(x) \sqrt{1+f'(x)^2} dx$$ This matches the given formula for the area of a surface of revolution, which completes the proof.

Key concepts

Calculus

Calculus might seem daunting at first, but it's really just a branch of mathematics that helps us study change and motion. Imagine you're driving a car, and you want to know how fast you're going - calculus helps you figure that out. But it also tells you how your speed changes over time, which is essential when considering more complex scenarios like the paths of planets or the growth of a population. And when it comes to finding the surface area of shapes that are created by rotating graphs around an axis - like the problem we're looking at - calculus is the tool we use to get there.

Differential Calculus

Differential calculus is all about instant change - think of it as the mathematics of motion. Whenever you hear 'rate of change', 'slope', or 'derivative', you're in the realm of differential calculus. In our exercise, when we discuss the function's derivative, f'(x), we're referring to how fast the function f is changing at each point. This concept is crucial because it allows us to calculate the tiny changes that occur when we rotate a curve around an axis, which we need for determining the surface area of the resulting shape.

Integrals

Where differential calculus involves rates and small pieces, integrals are about putting those pieces together to find whole sizes or quantities. Think of it like this: if you were building a brick wall, differential calculus would help you figure out the volume of each brick, while integrals would help you calculate the volume of the entire wall. In the context of our problem, we use an integral to 'sum up' all the tiny strips of area on the surface of our shape to get the total surface area.

Arc Length

Arc length describes the 'distance' along a curve, and calculating it can be a bit trickier than finding the length of a straight line. Imagine unrolling a curved road onto a straight path and measuring that distance - that's what arc length is about. Differential calculus comes into play here because to find an arc length, we break down the curve into infinitesimally small segments, find the length of each part (that's where ds comes from in our solution), and then add them all up using an integral to get the total distance. This step is essential for figuring out the surface area of the shape we get from rotating the curve, as the arc length of the curve directly impacts the area calculation.