Problem 22 - Thomas Calculus

Question

Find the areas of the surfaces generated by revolving the curves in Exercises \(13-23\) 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 (Hint: 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 area of the surface is \( 4\pi \).

Step 1: Understand the Problem

We need to find the surface area of the solid generated by revolving the curve \( y = \frac{1}{3}(x^2+2)^{3/2} \) from \( x = 0 \) to \( x = \sqrt{2} \) around the \( y \)-axis. We'll use the formula for the surface area of a surface of revolution around the \( y \)-axis, which involves finding an expression for the arc length element \( ds \).

Step 2: Find the Derivative

To find \( ds \), we first need the derivative of \( y \) with respect to \( x \): \[ y = \frac{1}{3}(x^2+2)^{3/2} \].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: Express \( ds \) in Terms of \( dx \)

The differential arc length \( ds \) is given by: \[ ds = \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \].Substitute \( \frac{dy}{dx} = x(x^2+2)^{1/2} \) into the formula:\[ ds = \sqrt{1 + \left(x(x^2+2)^{1/2}\right)^2} \, dx = \sqrt{1 + x^2(x^2+2)} \, dx = \sqrt{x^4 + 2x^2 + 1} \, dx \].

Step 4: Simplify the Expression for \( ds \)

Notice that:\[ x^4 + 2x^2 + 1 = (x^2 + 1)^2 \].So, \[ ds = (x^2 + 1) \, dx \].

Step 5: Set up the Surface Area Integral

The surface area \( S \) of a surface of revolution around the \( y \)-axis is given by: \[ S = \int 2\pi x \cdot ds \].Substitute \( ds = (x^2 + 1) \, dx \):\[ S = \int_{0}^{\sqrt{2}} 2\pi x (x^2 + 1) \, dx \].

Step 6: Evaluate the Integral

Calculate the integral:\[ \int 2\pi x(x^2 + 1) \, dx = 2\pi \left[ \int (x^3 + x) \, dx \right] \].The integral becomes:\[ 2\pi \left[ \frac{x^4}{4} + \frac{x^2}{2} \right]_{0}^{\sqrt{2}} = 2\pi \left[ \frac{(\sqrt{2})^4}{4} + \frac{(\sqrt{2})^2}{2} \right] \].Simplify:\[ 2\pi \times \left[ \frac{4}{4} + 1 \right] = 2\pi \times 2 = 4\pi \].

Step 7: Conclusion

The surface area of the solid formed by revolving the given curve about the \( y \)-axis is \( 4\pi \).

Key Concepts

Calculus

Calculus is a branch of mathematics that studies how things change. It provides tools and techniques to analyze changes in quantities, especially focusing on rates of change and accumulation.
It splits into two main areas: differential calculus and integral calculus.
Here, we are specifically interested in using calculus to solve problems involving curves and surfaces.
Differential calculus helps us understand how a function changes at any point by using derivatives, while integral calculus allows us to calculate the accumulation of quantities, such as areas and volumes.
This problem involves finding a surface area generated by revolving a curve, which requires both differential and integral calculus. By understanding how these curvatures behave (using derivatives) and accumulating them (using integrals), we can find the total surface area.

Arc Length

Arc length is the distance along a curve between two points. Calculating the arc length of a curve involves slicing it into small segments, approximating each piece with straight lines, and adding them up.
This geometric calculation connects directly with calculus when dealing with smooth curves.
For an arc length of a curve given by a function, the formula is: \[ ds = \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]This expression gives a small segment, or "differential element," of the arc length representing each tiny piece of the curve. By integrating this from one end of the interval to the other, we accumulate these lengths to find the total arc.
In this exercise, we use the arc length element \( ds \) to help find the surface area revolving around the y-axis, which involves stretching these arc lengths into spaces on a surface.

Integral Calculus

Integral calculus focuses on the idea of accumulation of quantities and calculating total size, length, area, or volume.
This exercise involves finding the surface area, which is a perfect application of integral calculus.
Surface area calculations use integrals to sum up all tiny pieces of the surface. Here's the general idea for surfaces of revolution:

• First, define the curve and find the expression for the arc length \( ds \).
• The integral for surface areas is similar to adding up lengths but modified to stretch those lengths around an axis, forming what can be visualized as ribbons wrapping around a shape.

The resulting integral form is:\[ S = \int 2\pi x \, ds \]In this integral, the factor \( 2\pi x \) comes from the circle equation, linking calculus to geometric rotation. For this problem, we calculated:\[ S = \int_{0}^{\sqrt{2}} 2\pi x (x^2 + 1) \, dx \]This represents gathering all those small surface areas produced by revolving the arc length around the y-axis, effectively using integration to accumulate these incremental areas into a total surface area.

Chain Rule

The chain rule is a fundamental concept in differential calculus. It helps find the derivative of a composition of functions.
When dealing with more complex functions, simply taking the derivative of each part separately doesn't work.
Instead, the chain rule provides a way to handle the complexities by considering how an inside function affects a larger outside one.In this problem, it was crucial to determining the derivative of the function \[ y = \frac{1}{3}(x^2+2)^{3/2}\]Here, there are layers: the inside function is \( x^2 + 2 \), and the outside function is raising to the power of \( \frac{3}{2} \). To find the derivative using the chain rule:

• Differentiate the outer function: multiply by \( \frac{3}{2} \) and reduce the power by 1.
• Multiply this by the derivative of the inside function: \( 2x \).

Using the chain rule: \[ \frac{dy}{dx} = \frac{1}{3} \times \frac{3}{2}(x^2+2)^{1/2} \times 2x = x(x^2+2)^{1/2} \]This gives us the derivative needed to plug into the arc length formula, enriching our steps towards solving the main surface area problem.