Problem 22 - Thomas Calculus in SI Units

Question

Find the areas of the surfaces generated by revolving the curves about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. \(y=(1 / 3)\left(x^{2}+2\right)^{3 / 2}, \quad 0 \leq x \leq \sqrt{2} ; \quad y\) -axis \(\quad\) (Hint: \(\quad\) Express \(d s=\sqrt{d x^{2}+d y^{2}}\) in terms of \(d x,\) and evaluate the integral \(S=\int 2 \pi x d s\) with appropriate limits.)

Step-by-Step Solution

Answer

The surface area is \(4\pi\).

Step 1: Understanding the Revolved Surface

We need to find the surface area generated by revolving the given curve around the y-axis. The formula for the surface area S when revolving about the y-axis is: \[ S = \int 2\pi x \, dS \]where \( dS = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx \).

Step 2: Finding the Derivative

First, find \( \frac{dy}{dx} \) of the given function \( y = \frac{1}{3}(x^2 + 2)^{3/2} \). We use the chain rule: \[ \frac{dy}{dx} = \frac{1}{3} \times \frac{3}{2}(x^2 + 2)^{1/2} \times 2x = x(x^2 + 2)^{1/2} \]

Step 3: Calculating the Differential of Arc Length

Substitute \( \frac{dy}{dx} \) into the expression for \( dS \): \[ dS = \sqrt{1 + \left( x(x^2 + 2)^{1/2} \right)^2} dx = \sqrt{1 + x^2(x^2 + 2)} dx = \sqrt{1 + x^4 + 2x^2} dx = \sqrt{(x^2 + 1)^2} dx = (x^2 + 1) dx \]

Step 4: Setting Up the Integral

Substitute \( dS = (x^2 + 1) dx \) back into the integral for the surface area: \[ S = \int_{0}^{\sqrt{2}} 2\pi x(x^2 + 1) \, dx \]

Step 5: Evaluating the Integral

Compute the integral:\[ \int 2\pi x(x^2 + 1) \, dx = 2\pi \int (x^3 + x) \, dx \]Split the integral and evaluate:\[ = 2\pi \left( \frac{x^4}{4} + \frac{x^2}{2} \right) \Bigg|_0^{\sqrt{2}} \]\[ = 2\pi \left[ \frac{(\sqrt{2})^4}{4} + \frac{(\sqrt{2})^2}{2} \right] \]\[ = 2\pi \left[ \frac{4}{4} + \frac{2}{2} \right] \]\[ = 2\pi (1 + 1) = 4\pi \]

Step 6: Final Result

The total surface area generated by revolving the curve \( y = \frac{1}{3}(x^2 + 2)^{3/2} \) around the y-axis is \( 4\pi \).

Key Concepts

Integral Calculus

Integral calculus is all about finding the total effect of something by summing up small pieces. In this exercise, we're focusing on the *Surface Area of Revolution*. Here, we use integrals to calculate the surface area formed when a curve is revolved around an axis. When revolving a curve, the formula for surface area involves an integral, because the calculation needs to account for the continuously changing shape of the curve. For revolution around the y-axis, as in this exercise, the formula is: - \[ S = \int 2\pi x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] Breaking this down:- The \(2\pi x\) represents the circumference of a small slice parallel to the y-axis.- \(dS\), the arc differential, adds the complexity of the curve into our calculation.Understanding how these pieces come together in an integral is key to solving many real-world problems involving curves, like finding the area of a surface, the volume of a solid, or the length of a curve.

Arc Length

Arc length is the distance along a curve between two points. In our problem, we're using the arc length to help determine the surface area of revolution. The differential arc length, \(dS\), covers the small length of a curve segment. When you add up these small segments, you find the total length of the arc. The formula for arc length combines both the x component and the y component:- \[dS = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \cdot dx\] This expression provides a way to factor in how your curve changes as both x and y change. The \(\frac{dy}{dx}\) is the slope of the curve at any point x, giving an insight into how steep or flat a segment of the curve is. When working with curves in calculus, remember:- Calculate \(\frac{dy}{dx}\) accurately as it heavily impacts your final calculations.- Integrating \(dS\) over an interval gives you the overall arc length, vital for accurately determining the characteristics of a curve's surface area.

Chain Rule

The chain rule is an essential tool in calculus, especially for handling problems involving composite functions. It shines through in our problem as we calculated the derivative \(\frac{dy}{dx}\). Understanding it well helps simplify complicated functions into easier parts. To use the chain rule, you must recognize when a function is composed of other functions. In our exercise, the function \(y = \frac{1}{3}(x^2 + 2)^{3/2}\) was the composite of the inner function \(u = x^2 + 2\) and the outer function \(f(u) = u^{3/2}\).Here's how the chain rule applies:- **Identify the inner and outer functions**: - Inner: \(u = x^2 + 2\) - Outer: \(f(u) = u^{3/2}\)- **Differentiate the outer function with respect to the inner function** and the inner function with respect to \(x\): - \(\frac{df}{du} = \frac{3}{2}u^{1/2}\) - \(\frac{du}{dx} = 2x\)- **Apply Chain Rule**: - \(\frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = \frac{3}{2}u^{1/2} \cdot 2x = x(x^2 + 2)^{1/2}\)Using the chain rule effectively allows a problem that initially looks daunting to become straightforward and solvable. Mastery of it unlocks otherwise difficult calculus problems.