Problem 27 - Calculus Single Variable

Question

Calculate the area \(S\) of the surface obtained when the graph of the given function is rotated about the \(x\) -axis. $$ f(x)=3 x^{1 / 2} \quad [7 / 4,4] $$

Step-by-Step Solution

Answer

The surface area is \(61\pi\).

Step 1: Understanding the Area of Revolution Formula

To find the surface area of the solid formed by rotating a function around the x-axis, we use the formula for the surface area of revolution: \[ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( f'(x) \right)^2} \, dx \].Here, \(f(x) = 3x^{1/2}\) and the interval is \([7/4, 4]\).

Step 2: Calculate the Derivative of the Function

We need to find \( f'(x) \). Since \( f(x) = 3x^{1/2} = 3x^{0.5} \), we use the power rule for differentiation:\[ f'(x) = 3 \cdot 0.5 x^{-0.5} = \frac{3}{2\sqrt{x}} \].

Step 3: Set Up the Integral for Surface Area

Substitute \(f(x)\) and \(f'(x)\) into the surface area formula:\[ S = 2\pi \int_{7/4}^{4} 3x^{1/2} \sqrt{1 + \left( \frac{3}{2\sqrt{x}} \right)^2} \ dx \].This simplifies to:\[ S = 6\pi \int_{7/4}^{4} x^{1/2} \sqrt{1 + \frac{9}{4x}} \ dx \].

Step 4: Simplify the Inside of the Square Root

Calculate the expression inside the square root:\[ 1 + \frac{9}{4x} = \frac{4x + 9}{4x} \].Thus,\[ \sqrt{1 + \frac{9}{4x}} = \sqrt{\frac{4x + 9}{4x}} = \frac{\sqrt{4x + 9}}{\sqrt{4x}} \].

Step 5: Substitute Back and Prepare to Integrate

Substitute back into the integral formula:\[ S = 6\pi \int_{7/4}^{4} x^{1/2} \cdot \frac{\sqrt{4x + 9}}{\sqrt{4x}} \ dx \],which simplifies to:\[ S = 6\pi \int_{7/4}^{4} \frac{x^{1/2} \sqrt{4x + 9}}{2 \sqrt{x}} \ dx \],giving\[ S = 3\pi \int_{7/4}^{4} \sqrt{4x + 9} \ dx \].

Step 6: Solve the Integral

To compute the integral \( \int_{7/4}^{4} \sqrt{4x + 9} \ dx \), perform a substitution: Let \( u = 4x + 9 \), then \( du = 4 \, dx \), meaning \( dx = \frac{1}{4} du \).Change the limits of integration: when \( x = 7/4 \), \( u = 16 \); when \( x = 4 \), \( u = 25 \).The integral becomes:\[ S = 3\pi \int_{16}^{25} \frac{1}{4} u^{1/2} \ du \].

Step 7: Integrate and Solve

Integrate the expression:\[ \int u^{1/2} \ du = \frac{2}{3} u^{3/2} \].Apply this to the modified limits:\[ = 3\pi \times \frac{1}{4} \left[ \frac{2}{3} u^{3/2} \right]_{16}^{25} \], which equals:\[ = \frac{\pi}{2} \left[ 2 \times (25^{3/2} - 16^{3/2}) \right] \].

Step 8: Calculate Final Values

Compute the values:\[ 25^{3/2} = 125 \text{ and } 16^{3/2} = 64 \].Thus, the surface area is:\[ S = \frac{\pi}{2} \times (250 - 128) = \frac{\pi}{2} \times 122 = 61\pi \].

Key Concepts

Integral Calculation

Integral calculation is a fundamental concept in calculus that helps us find quantities such as areas and volumes. In this context, when computing the area of a surface of revolution, we set up and solve integrals to determine the surface area. The surface area of a solid of revolution is calculated using the integral formula:

• \[ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left( f'(x) \right)^2} \, dx \]

This integral formula takes into account both the function describing the shape being revolved and its derivative. The integral bounds \([a, b]\) indicate the interval over which the calculation is performed.
To solve integrals effectively, one often needs to simplify the integrand, especially when it includes complex expressions such as square roots. This might involve algebraic manipulations or substitutions to make the integral easier to evaluate.

Differentiation

Differentiation is a process that involves finding the derivative of a function. The derivative measures how a function changes as its input changes, effectively describing its rate of change at any given point. When dealing with the surface area of revolution, knowing the derivative \(f'(x)\) is crucial because it is part of the formula used to find the surface area.
For the given function \(f(x) = 3x^{1/2}\), we apply the power rule to differentiate it:

• Convert the expression to \(3x^{0.5}\).
• Use the power rule given by \(nx^{n-1}\) to differentiate, resulting in \(f'(x) = 3 \cdot 0.5 x^{-0.5} = \frac{3}{2\sqrt{x}}\).

This derivative is then utilized in the integral to help determine the effect of revolving the function around the x-axis.

Power Rule

The power rule is a basic but essential concept in calculus. It provides a straightforward way to differentiate power functions. Generally, if you have a function \(x^n\), the power rule tells us that its derivative is \(nx^{n-1}\). This rule is not only useful for differentiation but also appears in integral calculations as it helps recognize antiderivatives.
In the exercise, the function \(f(x) = 3x^{1/2}\) requires differentiation using the power rule:

• Re-write the expression as \(3x^{0.5}\).
• Apply the rule: differentiate to obtain \(f'(x) = \frac{3}{2\sqrt{x}}\).

This simple method of differentiation streamlines the process of evaluating changes in the function, crucial for calculating areas or volumes.

Definite Integral

The definite integral is a key concept that helps us compute the area under a curve or, in this case, the surface area formed by revolving a curve around an axis. It is indicated by integral signs with upper and lower limits, \(\int_{a}^{b}\), representing the interval over which the function is integrated.
In our problem, once the function and its derivative are known, we insert these into the surface area formula:

• Substitute values and simplify: \[ S = 3\pi \int_{7/4}^{4} \sqrt{4x + 9} \, dx \].

Breaking this integral into manageable parts may involve substitutions, like letting \(u = 4x + 9\), to simplify evaluation. Adjust the limits accordingly from \([7/4, 4]\) to \([16, 25]\) for \(u\). Once integrals are computed, these lead us to the solution, providing an exact area value. This demonstrates how the definite integral is a powerful tool for precise calculation in real-life applications.