Problem 36 - Essential Calculus

Question

In Exercises \(35-38\), set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the \(x\) -axis. $$ y=2 \sqrt{x}, \quad 4 \leq x \leq 9 $$

Step-by-Step Solution

Answer

To get the surface area, compute the integral set up in the second step. As stated, it often involves complex calculations that are typically carried out with a calculator. The resolved value is the final surface area.

Step 1: Compute the derivative

The given function is \(y = 2 \sqrt{x}\) which can be rewritten as \(y = 2x^{1/2}\). The derivative of this function is: \[y^{'} = \frac{d y}{d x} = 2*\frac{1}{2}*x^{-1/2} = x^{-1/2}.\]

Step 2: Set up the integral for the surface area

Substituting the function and its derivative into the surface area formula and between the limits of x give: \[A = 2\pi \int_{4}^{9} 2\sqrt{x} \sqrt{1+(x^{-1/2})^2} dx.\]

Step 3: Evaluate the integral

Evaluate the integral either analytically or with calculator to get the value of the surface area. The integral is most of times computed with a calculator due to its complexity.

Key Concepts

Surface Area

The concept of surface area is crucial when dealing with shapes in geometry and calculus. It represents the total area that the surface of an object occupies.

In calculus, calculating the surface area of a shape involves integrating over a particular region. This requires setting up a definite integral, especially when dealing with shapes formed by revolving a curve around an axis.

For example, when you need to find the surface area of a shape formed by revolving a curve around the x-axis, you start with the surface area formula:

• When the curve is given as a function y = f(x), the formula becomes:
  \[ A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + (f'(x))^2} \, dx \]
• Here, a and b are the limits over which you are integrating.
  The expression \( f'(x) \) is the derivative of the function \( f(x) \).

This setup allows for precise calculation of the surface area generated by the revolving curve.

Revolving Curve

Revolving a curve around an axis is a fascinating concept in calculus that helps us visualize and calculate three-dimensional shapes from two-dimensional functions.

To generate a solid through this method, consider a given curve, such as \( y = 2 \sqrt{x} \) in this exercise. When revolving this curve around the x-axis between the limits \( x = 4 \) and \( x = 9 \), it creates a specific shape known as a surface of revolution.

These surfaces have applications in many fields, including physics and engineering, where they assist in designing objects with rotational symmetry, like pipes and containers. By setting up integrals, we can precisely calculate characteristics such as surface area and volume. This exercise focuses on finding the surface area when the curve \( y = 2 \sqrt{x} \) revolves, using integration techniques.

This method fundamentally enhances our understanding of how two-dimensional curves can form complex three-dimensional structures.

Derivative

Derivatives are a core component of calculus, representing how a function changes as its input changes.

In this exercise, we start with the function \( y = 2 \sqrt{x} \), which represents the curve to be revolved. To calculate the surface area of the resulting shape, we first find the derivative of the function.

The derivative of \( y = 2 \sqrt{x} \) is calculated by rewriting the function as \( y = 2x^{1/2} \) and using the power rule for differentiation. This rule tells us that the derivative of \( x^n \) is \( nx^{n-1} \).

Applying this, we get:

• \( y' = \frac{d}{dx} [2x^{1/2}] = 1 \cdot x^{-1/2} \).

This derivative is essential for the surface area formula, as it helps to account for the rate at which the curve's slope changes, which influences the geometry of the revolving surface.

Integral Evaluation

Integral evaluation is a crucial step in finding the actual surface area, as outlined by mathematical formulas.

It involves setting up and solving a definite integral, where the limits are provided by the exercise. For this exercise, these limits are \( x = 4 \) to \( x = 9 \).

The integrand for the surface area includes both the original function \( 2\sqrt{x} \) and its derivative \( x^{-1/2} \), which is inserted into the integral formula:

• \[ A = 2\pi \int_{4}^{9} 2\sqrt{x} \sqrt{1 + (x^{-1/2})^2} \, dx \]

Solving this integral, often done with a calculator due to its complexity, gives the numerical value of the surface area created by the revolving curve.

Mastering integral evaluation in this context involves understanding how to manipulate the integral and apply either analytical techniques or computational tools to achieve a solution, illustrating the interplay between calculus theory and practical application.