Problem 19 - Calculus Early Transcendentals

Question

Find the area of the surface generated when the given curve is revolved about the \(y\) -axis. The part of the curve \(y=4 x-1\) between the points (1,3) and (4,15)

Step-by-Step Solution

Answer

Answer: The area of the surface generated is \(240\pi\sqrt{17}\).

Step 1: Find the inverse function x(y)

To find the inverse function \(x(y)\), we need to solve the equation \(y = 4x - 1\) for \(x\). We can do this by adding 1 to both sides and then dividing by 4: \(y + 1 = 4x\) \(x = \frac{y + 1}{4}\) So, \(x(y) = \frac{y + 1}{4}\).

Step 2: Find the derivative y'(x)

Next, we need to find the derivative of \(y\) with respect to \(x\). Since we have \(y = 4x - 1\), we can find \(y'(x)\) by taking the derivative with respect to \(x\): \(y'(x) = \frac{d}{dx}(4x - 1) = 4\)

Step 3: Plug x(y) and y'(x) into the formula

Now we can plug \(x(y)\) and \(y'(x)\) into the formula for the area of the surface of revolution: \(A = 2\pi\int_{3}^{15}\frac{y + 1}{4}\sqrt{1 + (4)^2} dy\)

Step 4: Evaluate the integral

Now let's evaluate the integral. We have: \(A = 2\pi\int_{3}^{15}\frac{y + 1}{4}\sqrt{17} dy = \frac{2\pi\sqrt{17}}{4}\int_{3}^{15}(y + 1) dy\) To evaluate the integral, we can use the power rule for integration: \(\int_{3}^{15}(y + 1) dy = \frac{1}{2}(y + 1)^2 \Big|_{3}^{15}\) Evaluating the integral at the limits of integration, we have: \(\frac{1}{2}(15 + 1)^2 - \frac{1}{2}(3 + 1)^2 =\frac{1}{2}(256 - 16)\)

Step 5: Calculate the area

Finally, let's plug the value of the integral back into the formula for the area: \(A = \frac{2\pi\sqrt{17}}{4}\left(\frac{1}{2}(256 - 16)\right) = \pi\sqrt{17}(240)\) Therefore, the area of the surface generated when the given curve is revolved about the y-axis is \(240\pi\sqrt{17}\).

Key Concepts

Inverse Function

An inverse function essentially reverses the operation done by the original function. In this exercise, we start with the function \( y = 4x - 1 \), and our goal is to express \( x \) in terms of \( y \).

To find the inverse, we need to isolate \( x \). We add 1 to both sides to get \( y + 1 = 4x \) and then divide everything by 4. The inverse function is \( x(y) = \frac{y + 1}{4} \).

This transformation is crucial because, to find the surface area of revolution around the \( y \)-axis, we need \( x \) in terms of \( y \). By knowing the inverse, we can integrate with respect to \( y \), which is necessary for this calculation.

Definite Integral

A definite integral is a powerful tool used to calculate values across a certain interval, such as areas under curves or, in our case, the area of a surface.

In the surface area problem, the integral expressions are evaluated from \( y = 3 \) to \( y = 15 \). These limits correspond to the vertical bounds derived from the given curve \( y = 4x - 1 \).

The expression to solve is:

• \( \int_{3}^{15} \frac{y + 1}{4}\sqrt{1 + (4)^2} \ dy \).

The definite integral, when evaluated over this range, helps us find the precise value needed for the surface area of revolution.

Power Rule in Integration

The power rule for integration is a fundamental principle that allows you to integrate functions of the form \( x^n \). Essentially, it states that to integrate \( x^n \), you add one to the exponent and then divide by the new exponent. For any function \( y^n \), the integral becomes:

• \( \int y^n \ dy = \frac{1}{n+1} y^{n+1} + C \).

In our exercise's integration step, the problem includes \( y + 1 \) as the expression inside the integral. When the power rule is applied, the integral becomes:

• \( \int (y + 1) \ dy = \frac{1}{2}(y + 1)^2 \)

Evaluating this expression at the specified limits helps us to form the final area calculation.

Surface of Revolution Formula

The concept of a surface of revolution revolves around rotating a curve about an axis to form a 3-Dimensional shape. To find the area of such a surface, mathematicians use a specialized formula.

When revolving a curve about the \( y \)-axis, the formula becomes:

• \( A = 2\pi \int_{a}^{b} x(y)\sqrt{1 + (y'(x))^2} \, dy \)

This formula incorporates the inverse function \( x(y) \) and the derivative \( y'(x) \).
In the exercise, the derivative does not vary \( y'(x) = 4 \), leading to simplification in the square root term.

Finally, the integral resolves to give the surface area of the revolution, delivering the area \( 240\pi\sqrt{17} \).
This formula elegantly connects derivatives, integrals, and inverse functions, all essential components for solving problems of this nature.