Problem 22 - Thomas Calculus

Question

Find the areas of the surfaces generated by revolving the curves in Exercises \(13-22\) 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 \((\text {Hint} : \text { Express }\) \(d s=\sqrt{d x^{2}+d y^{2}}\) 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 \( S = \pi \left( \frac{16 - 4\sqrt{2}}{3} \right) \).

Step 1: Set up for the Surface Area Integral

We want to find the surface area of the curve revolved around the y-axis. The formula for the surface area of a curve revolved around the y-axis is given by:\[ S = \int 2\pi x \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]where \( x \) is the radius of revolution and \( \frac{dy}{dx} \) is the derivative of \( y \) with respect to \( x \). Our function is \( y = \frac{1}{3}(x^2 + 2)^{3/2} \), so we need to find \( \frac{dy}{dx} \).

Step 2: Differentiate y with respect to x

To find \( \frac{dy}{dx} \), we differentiate the function \( y = \frac{1}{3}(x^2 + 2)^{3/2} \):Use the chain rule. Let \( u = x^2 + 2 \). Then, \( y = \frac{1}{3}u^{3/2} \), and we differentiate:\[ \frac{dy}{du} = \frac{1}{3} \cdot \frac{3}{2} u^{1/2} = \frac{1}{2} u^{1/2} \]\[ \frac{du}{dx} = 2x \]Thus, \( \frac{dy}{dx} = \frac{1}{2} (x^2 + 2)^{1/2} \times 2x = x(x^2 + 2)^{1/2} \).

Step 3: Substitute into Surface Area Integral

Now substitute \( \frac{dy}{dx} = x(x^2 + 2)^{1/2} \) into the surface area integral:\[ S = \int_0^{\sqrt{2}} 2\pi x \sqrt{1 + \left( x(x^2+2)^{1/2} \right)^2} \, dx \]Simplify the integrand:\[ 1 + (x(x^2+2)^{1/2})^2 = 1 + x^2(x^2 + 2) = x^2 + 1 \times x^2 + x^2 \times 2 = x^2(x^2 + 2) = x^2(x^2 + 2) \] Then,\[ \sqrt{1 + x^2(x^2+2)} = \sqrt{x^2+2} \]

Step 4: Evaluate the Integral

Now the integral simplifies to:\[ S = \int_0^{\sqrt{2}} 2\pi x (x^2 + 2)^{1/2} \, dx \]To evaluate the integral, use substitution. Let \( u = x^2 + 2 \), then \( du = 2x \, dx \), or \( x \, dx = \frac{1}{2} du \).When \( x = 0, u = 2 \) and when \( x = \sqrt{2}, u = 4 \).The bounds change accordingly:\[ S = \pi \int_2^4 u^{1/2} \, du \]Integrating, we find:\[ S = \pi \left[ \frac{2}{3} u^{3/2} \right]_2^4 = \pi \left( \frac{2}{3} \times 4^{3/2} - \frac{2}{3} \times 2^{3/2} \right) \]\[ = \pi \left( \frac{2}{3} \times 8 - \frac{2}{3} \times 2\sqrt{2} \right) = \pi \left( \frac{16}{3} - \frac{4\sqrt{2}}{3} \right) \]

Step 5: Compute Final Surface Area Result

Simplify the expression obtained from the integral:\[ S = \pi \left( \frac{16}{3} - \frac{4\sqrt{2}}{3} \right) \]This computes to:\[ S = \frac{16\pi}{3} - \frac{4\pi\sqrt{2}}{3} \]The total surface area of the revolution about the y-axis is:\[ S = \pi \left( \frac{16 - 4\sqrt{2}}{3} \right) \]

Key Concepts

Integral Calculus

Integral calculus is a branch of calculus that focuses on finding the total accumulation of quantities. It's the counterpart to differential calculus, which instead deals with rates of change. In integral calculus, we solve problems related to calculating areas under curves, volumes of solids, and other quantities that can be seen as an accumulation.

To solve integrals, we use techniques such as substitution (changing variables for simpler computation) and integration by parts, which is particularly handy when dealing with products of functions inside integrals. In our original exercise, we calculate a surface area by setting up a specific integral that represents this accumulated surface.

• Surface integrals often include a factor related to the geometry of the problem, like a radius when dealing with rotations around an axis.
• Evaluating these integrals can involve substituting expressions that simplify into new variables to make the integral easier to solve.

Integrals are tools that allow us to make sense of complex shapes and changes, essential for applications in physics, engineering, and beyond.

Chain Rule

The chain rule is a fundamental technique in calculus for finding derivatives of composite functions. This rule states that if you have a function nested within another function, like a 'chain' of functions, you differentiate the outer function and multiply it by the derivative of the inner function.

In the problem at hand, to differentiate the function \( y = \frac{1}{3}(x^2 + 2)^{3/2} \), we see this composition. First, identify the outer function as \( u^{3/2} \) and the inner function as \( u = x^2 + 2 \).

• Start by differentiating the outer function with respect to the inner function: \( \frac{dy}{du} = \frac{1}{2}u^{1/2} \).
• Next, differentiate the inner function with respect to \( x \): \( \frac{du}{dx} = 2x \).
• Combine these using the chain rule to get: \( \frac{dy}{dx} = \frac{1}{2}u^{1/2} \cdot 2x \), resulting in \( x(x^2 + 2)^{1/2} \).

The chain rule is essential for navigating complex functions, especially in problems involving nested expressions like this one.

Surface Integration

Surface integration involves techniques to find the surface area of objects, especially those formed by revolving curves around an axis. This concept extends integral calculus to two dimensions when you calculate the 'skin' over a region.

In our exercise, we are dealing with the surface area generated by a curve around the y-axis. The formula essential for solving such a problem is:\[ S = \int 2\pi x \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \ dx \]Here, \( 2\pi x \) represents the circumference of the revolving circle created by a slice of the curve at a particular \( x \)-value.

• The term \( \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \) accounts for the 'stretch' of the curve, accounting both for the change in the y-direction (height) and x-direction (width).
• Evaluating this integral can simplify to more manageable algebra by substitution, allowing easier computation of areas involving complex functions.

Surface integration allows us to quantify the extent of surfaces, a crucial aspect of geometry and applications involving physical shapes and spaces.