Question

Find the area of the surface formed by revolving the graph of \(y=2 \sqrt{x}\) over the interval \([0,9]\) about the \(x\) -axis.

Short answer

The surface area is approximately 256.68 square units.

Step-by-step solution

Understanding the Problem

The goal is to find the surface area of the figure formed by revolving the curve \(y = 2 \sqrt{x}\) from \(x=0\) to \(x=9\) around the \(x\)-axis. This is done using the surface area of revolution formula, which is \(S = \int_a^b 2\pi y \sqrt{1 + (\frac{dy}{dx})^2} \,dx\) for a curve \(y = f(x)\).

Find the Derivative

First, we need to find the derivative \(\frac{dy}{dx}\) of the function \(y = 2\sqrt{x}\). Using the power rule, the derivative is \(\frac{dy}{dx} = \frac{1}{\sqrt{x}}\).

Set Up the Integral for Surface Area

Substitute \(y = 2\sqrt{x}\) and \(\frac{dy}{dx} = \frac{1}{\sqrt{x}}\) into the surface area formula: \[ S = \int_0^9 2\pi (2\sqrt{x}) \sqrt{1 + \left(\frac{1}{\sqrt{x}}\right)^2} \,dx \]. Simplify the expression under the square root: \(1 + \left(\frac{1}{\sqrt{x}}\right)^2 = 1 + \frac{1}{x}\).

Simplify the Integral

Simplify the square root to get \(\sqrt{1 + \frac{1}{x}} = \sqrt{\frac{x+1}{x}} = \frac{\sqrt{x+1}}{\sqrt{x}}\). Substituting this into the integral gives: \[ S = \int_0^9 4\pi \sqrt{x} \cdot \frac{\sqrt{x+1}}{\sqrt{x}} \, dx = \int_0^9 4\pi \sqrt{x+1} \, dx \].

Evaluate the Integral

Use substitution: let \(u = x + 1\), then \(du = dx\) and \(\sqrt{x+1} = \sqrt{u}\). The limits become \(x=0\) is \(u=1\) and \(x=9\) is \(u=10\). The integral becomes: \[ \int_1^{10} 4\pi \sqrt{u} \, du \]. Evaluate the integral: \[ 4\pi \int_1^{10} u^{1/2} \, du = 4\pi \left[ \frac{2}{3} u^{3/2} \right]_1^{10} \].

Calculate the Definite Integral

Compute the definite integral: \[ 4\pi \left[ \frac{2}{3} (10^{3/2}) - \frac{2}{3} (1^{3/2}) \right] = 4\pi \left[ \frac{2}{3} (31.622) - \frac{2}{3} (1) \right] \]. This simplifies to \[ 4\pi \left[ 21.081 - \frac{2}{3} \right] = 4\pi \times 20.414 \].

Calculate the Surface Area

Finish the computation: \[ S = 4\pi \times 20.414 \approx 256.68 \]. Therefore, the surface area of the surface formed by revolving the curve around the \(x\)-axis is approximately 256.68 square units.

Key concepts

Calculus in Surface Area Problems

Calculus plays a crucial role in understanding and solving surface area of revolution problems. At its core, calculus is the mathematical study of change and motion. It allows us to calculate various types of areas, slopes, and rates of change in the context of functions.
When dealing with curves and their revolutions in geometry, calculus steps in with its powerful tools to help compute those intricate shapes. One of the key concepts in calculus used here is the idea of integration, which allows us to add up infinitely many tiny portions to find an overall measurement like a surface area.

• Calculus provides tools to understand and calculate dynamic changes in mathematics.
• It helps in determining areas, volumes, and other quantities related to geometric shapes.
• Integration, a key feature of calculus, is essential in finding areas under curves or surfaces.

Integration and its Role

Integration is like your mathematical toolbox for accumulating quantity over intervals—think of it as sophisticated summing. In the context of surface area of revolution, integration takes on a special significance. It helps us to sum up the infinite number of tiny slices (called infinitesimals) over a curve to find the complete area.
This process involves the integral function, which acts as a grand totalizer, accommodating the rise and fall of the function’s curve over an interval.

• Integration sums up parts to determine a whole.
• It deals with continuous functions over specific intervals.
• Using definite integrals, we can precisely compute areas like that of a surface of revolution.

Derivatives Simplified

In calculus, derivatives are the mathematical way to describe how something changes. They're like the telltales of how a curve on a graph grows, resonates, or flattens out.
For our surface area problem, computing the derivative of the curve's equation, such as with the function \( y = 2\sqrt{x} \), gives insight into how steep the curve is at any given point. This changing gradient is a crucial component when setting up our integral to determine the surface area.

• Derivatives inform us about the direction and steepness of curves.
• They play a key role in various formulas involving rates of change.
• In surface area calculations, the derivative helps address how the area grows with each incremental slice.

Definite Integrals in Practice

A definite integral is a math concept that makes it possible to calculate exact areas or accumulated quantities over a specific interval. Unlike indefinite integrals, which provide a general formula, definite integrals give you a precise numerical value.
In our example, after finding the original function and its derivative, we used the definite integral to calculate the surface area of the revolution by integrating over the interval \( [0, 9] \). We adjusted the limits to account for substitution, turning initial limits into \( [1, 10] \).

• Definite integrals provide specific numerical values over an interval.
• They require defined upper and lower limits.
• It’s essential for precise calculations like the surface area of revolved surfaces.