Problem 18 - Calculus Early Transcendentals

Question

Find the area of the surface generated when the given curve is revolved about the \(y\) -axis. $$y=\frac{x^{2}}{4}, \text { for } 2 \leq x \leq 4$$

Step-by-Step Solution

Answer

The area of the surface generated is \(\frac{32\pi}{15}(5 - 2\sqrt{2})\).

Step 1: Find the derivative of the function with respect to x

To find the derivative \(\frac{dy}{dx}\), we will differentiate the function \(y=\frac{x^2}{4}\) with respect to \(x\): $$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{x^2}{4}\right)=\frac{1}{2}x$$

Step 2: Substitute the function and its derivative into the Surface Area of Revolution Formula

Now, we will substitute the function \(y=\frac{x^2}{4}\) and its derivative \(\frac{dy}{dx} = \frac{1}{2}x\) into the Surface Area of Revolution Formula: $$A=2\pi\int_{a}^{b} x \sqrt{1+\left(\frac{1}{2}x\right)^2} dy$$

Step 3: Rewrite the integral in terms of y

Our integral is written in terms of \(dy\), but our function and its derivative are in terms of \(x\) so we need to rewrite our integral in terms of \(y\). We can do this by rewriting \(x\) as a function of \(y\), using the original function \(y=\frac{x^2}{4}\). From that, $$x=2\sqrt{y}$$ Now we will also need to find an equivalent expression for \(dx\). From the rewritten equation, we have that $$x=2\sqrt{y} \Rightarrow dx=\frac{1}{\sqrt{y}}dy$$

Step 4: Substitute the expressions for x and dx into the integral

Now, substitute the expressions for \(x\) and \(dx\) into the formula: $$A=2\pi\int_{a}^{b} (2\sqrt{y}) \sqrt{1+\left(\frac{(2\sqrt{y})}{2} \right)^2} \frac{1}{\sqrt{y}} dy$$

Step 5: Simplify the integral and change the limits

First, simplify the integral: $$A=2\pi\int_{a}^{b} 2\sqrt{y} \sqrt{1+y} dy$$ We need to change the limits of integration from values of \(x\) to values of \(y\). From our original function, we can find the corresponding values of \(y\) for the given range of \(x\): When \(x=2\), \(y=\frac{(2)^2}{4}=1\) When \(x=4\), \(y=\frac{(4)^2}{4}=4\) The limits for \(y\) are now \(1 \leq y \leq 4\). Plug these as the limits of the integral: $$A=2\pi\int_{1}^{4} 2\sqrt{y} \sqrt{1+y} dy$$

Step 6: Evaluate the integral

Evaluate the integral: $$A=2\pi\int_{1}^{4} 2\sqrt{y} \sqrt{1+y} dy = 8\pi\left[\frac{2}{15}(1+y)^{3/2}\right]_1^4$$ Now, calculate the final value: $$A=8\pi\left[\frac{2}{15}(1+4)^{3/2}-\frac{2}{15}(1+1)^{3/2}\right] = \frac{32\pi}{15}(5 - 2\sqrt{2})$$ The area of the surface generated when the given curve is revolved about the y-axis is \(\frac{32\pi}{15}(5 - 2\sqrt{2})\).

Key Concepts

Calculus

Calculus is a branch of mathematics that helps us understand rates of change and the accumulation of quantities. It consists of two main components: differentiation and integration. Each provides tools to solve various problems such as finding slopes, areas, or volumes. Calculus allows us to translate real-world phenomena into mathematical models, which can be analyzed and solved. In the context of solving for the surface area of revolution, understanding how to apply calculus concepts enables us to calculate the area generated by revolving a curve around an axis.

Integration

Integration is a fundamental technique in calculus used to find areas under curves and accumulate quantities like distance or mass. In the problem we're tackling, integration is used to compute the surface area of revolution. The basic idea is to sum up, or integrate, an infinite number of infinitesimally thin, curved strips that approximate the surface area.

The formula we use for the surface area of revolution involves integrating the product of:

• The distance from the axis of rotation, which is represented by the function, and
• The length of the curve's arc, derived from the derivative of the function.

This turns a complex shape into a solvable integral expression.

Differentiation

Differentiation is the process of finding the derivative of a function, which tells us how the function is changing at any point. It's crucial in calculating tangents, velocities, and, in our exercise, arc lengths. For the surface area of revolution, differentiation helps in determining the rate at which the function value changes, which directly contributes to the element of arc length needed in the surface area formula.

In our example, we find the derivative of the function \(y=\frac{x^2}{4}\) to be \(\frac{1}{2}x\). This step is pivotal since the derivative informs us how steep or flat the curve is, affecting the surface generated when the curve is rotated around an axis.

Definite Integrals

Definite integrals are used to calculate the accumulation of quantities within a specific range. In the context of this exercise, definite integrals enable us to quantify the entire surface area of the shape generated from rotating a curve around an axis. The limits of the integral define the starting and ending points of this region.

In the given problem, the definite integral calculated is from the lower limit where \(x = 2\) to the upper limit at \(x = 4\). Hence, these values are translated into values of \(y\): from \(y = 1\) to \(y = 4\), which are then used in the integration bounds for solving the surface area.