Problem 37 - Calculus Early Transcendentals

Question

Let \(f\) be a nonnegative function with a continuous first derivative on \([a, b]\) and suppose that \(g(x)=c f(x)\) and \(h(x)=f(c x),\) where \(c>0 .\) When the curve \(y=f(x)\) on \([a, b]\) is revolved about the \(x\) -axis, the area of the resulting surface is \(A .\) Evaluate the following integrals in terms of \(A\) and \(c\). $$\text { a. } \int_{a}^{b} 2 \pi g(x) \sqrt{c^{2}+g^{\prime}(x)^{2}} d x \quad \text { b. } \int_{a / c}^{b / c} 2 \pi h(x) \sqrt{c^{2}+h^{\prime}(x)^{2}} d x$$

Step-by-Step Solution

Answer

Question: Determine the values of the following integrals in terms of A and c: (a) \(\int_{a}^{b} 2 \pi g(x) \sqrt{c^2+g'(x)^2} dx\) (b) \(\int_{a/c}^{b/c} 2 \pi h(x) \sqrt{c^2+h'(x)^2} dx\) Answer: (a) \(\int_{a}^{b} 2 \pi g(x) \sqrt{c^2+g'(x)^2} dx = 2\pi cA\) (b) \(\int_{a/c}^{b/c} 2 \pi h(x) \sqrt{c^2+h'(x)^2} dx = \frac{2\pi}{c}A\)

Step 1: Surface area of revolution formula

The formula for the surface area of revolution when rotating the curve \(y=f(x)\) around the x-axis between \(x=a\) and \(x=b\) is: $$A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + [f'(x)]^2} dx$$

Step 2: Find derivatives of g and h

We first need to find the first derivatives of functions \(g\) and \(h\). Let's start with \(g'\). Using the chain rule and given that \(g(x) = cf(x)\), we get: $$g'(x) = c\cdot f'(x)$$ Now let's find the derivative of \(h\). Given that \(h(x) = f(cx)\), we get: $$h'(x) = cf'(cx)$$

Step 3: Evaluate the integrals

(a) We need to evaluate the following integral in terms of \(A\) and \(c\): $$\int_{a}^{b} 2 \pi g(x) \sqrt{c^2+g'(x)^2} dx$$ Substitute \(g(x)\) and \(g'(x)\) into the integral: $$\int_{a}^{b} 2 \pi (cf(x)) \sqrt{c^2+(cf'(x))^2} dx$$ Now factor \(c\) out of the square root and then factor \(2\pi c\) out of the integral, which gives us: $$2\pi c \int_{a}^{b} f(x) \sqrt{1 + [f'(x)]^2} dx$$ Comparing this integral with the surface area of revolution formula for curve \(y=f(x)\), we can see that the integrand is exactly the same, which means that the integral is equal to \(A\). Therefore: $$\int_{a}^{b} 2 \pi g(x) \sqrt{c^2+g'(x)^2} dx = 2\pi cA$$ (b) Now, let's evaluate the following integral: $$\int_{a/c}^{b/c} 2 \pi h(x) \sqrt{c^2+h'(x)^2} dx$$ Substitute \(h(x)\) and \(h'(x)\) into the integral: $$\int_{a/c}^{b/c} 2 \pi (f(cx)) \sqrt{c^2+(cf'(cx))^2} dx$$ Now perform substitution \(u = cx\), \(du = c dx\), \(a/c=\frac{a}{c}\), and \(b/c=\frac{b}{c}\): $$\int_{\frac{a}{c}}^{\frac{b}{c}} 2 \pi f(u) \sqrt{c^2+c^2[f'(u)]^2} \frac{1}{c} du$$ Factor out \(2\pi\), cancel out c in the numerator and denominator, and notice that this again is the same as the surface area of revolution formula for curve \(y=f(x)\): $$\frac{2 \pi}{c} \int_{a/c}^{b/c} {f(u)} \sqrt{1 + [f'(u)]^2} du$$ Therefore, $$\int_{a/c}^{b/c} 2 \pi h(x) \sqrt{c^2+h'(x)^2} dx = \frac{2\pi}{c}A$$

Key Concepts

Integral Calculus

Integral calculus is an essential tool for understanding the concepts behind surface areas of revolution. It involves computing the integral of a function, which helps us find areas and volumes, among other things.
For the surface area of a shape formed by rotating a curve about an axis, we use the formula:

• \[ A = \int_{a}^{b} 2\pi f(x) \sqrt{1 + [f'(x)]^2} dx \]

This expression gives us the surface area of a curve \(y=f(x)\) when it is revolved around the \(x\)-axis.
Applying integral calculus allows us to evaluate this area accurately, making it a fundamental part of solving related problems.

Chain Rule

The chain rule is a crucial concept in calculus when dealing with composite functions, like \(g(x) = cf(x)\) and \(h(x) = f(cx)\).
It helps us find the derivative of a composite function by taking the derivative of the outer function and multiplying it by the derivative of the inner function.
For example:

• For \(g(x) = cf(x)\), we find \(g'(x) = c\cdot f'(x)\).
• For \(h(x) = f(cx)\), we find \(h'(x) = cf'(cx)\).

Using the chain rule allows precise calculation of derivatives necessary for further integral evaluation, which is vital in computing surface areas.

Derivative

A derivative represents the rate of change of a function. In our exercise, we needed the derivatives of \(g\) and \(h\), which involved applying the chain rule.
The derivative helps us understand how a function changes at any given point and is critical in forming the integral for surface area.

• Knowing \(f'(x)\) allows us to determine the term \(\sqrt{1+[f'(x)]^2}\) in the surface area formula.

Tracking these changes is crucial because it leads us to understand how the original curve contributes to the total surface area when revolved around an axis.

Function Transformation

Function transformation involves altering a function to create a new function. In this problem, we transformed \(f(x)\) into \(g(x) = cf(x)\) and \(h(x) = f(cx)\).
This manipulation helps in understanding how changing a function impacts its behavior in the context of surface areas of revolution.

• \(g(x) = cf(x)\) scales the height of the function by a constant \(c\).
• \(h(x) = f(cx)\) compresses or stretches the function horizontally depending on \(c\).

Being aware of these transformations allows you to adapt the surface area computation to different scenarios based on function changes and ensure accurate results for varied forms.