Problem 30 - 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. $$ f(x)=(x+1)^{1 / 2} \quad [1,11] $$

Step-by-Step Solution

Answer

The area of the surface is \( \frac{158\pi}{3} \).

Step 1: Understand the Problem

We need to find the area of the surface generated by rotating the curve \( y = f(x) = (x+1)^{1/2} \) around the \( x \)-axis from \( x = 1 \) to \( x = 11 \).

Step 2: Apply the Surface Area Formula

The formula for the surface area generated by rotating a curve \( y = f(x) \) from \( x = a \) to \( x = b \) around the \( x \)-axis is: \[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} \, dx \] In this case, \( f(x) = (x+1)^{1/2} \) and \( a = 1 \), \( b = 11 \).

Step 3: Compute the Derivative \( f'(x) \)

Find the derivative \( f'(x) \) of \( f(x) = (x+1)^{1/2} \). Using the chain rule, we have:\[ f'(x) = \frac{1}{2}(x+1)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x+1}} \]

Step 4: Set Up the Integral

Substitute \( f(x) \) and \( f'(x) \) into the surface area formula:\[ S = 2\pi \int_1^{11} (x+1)^{1/2} \sqrt{1 + \left(\frac{1}{2\sqrt{x+1}}\right)^2} \, dx \]Simplify the expression under the square root:\[ \sqrt{1 + \frac{1}{4(x+1)}} = \sqrt{\frac{4(x+1) + 1}{4(x+1)}} = \frac{\sqrt{4x+5}}{2\sqrt{x+1}} \]

Step 5: Simplify the Integral

Now the integral becomes:\[ S = 2\pi \int_1^{11} (x+1)^{1/2} \cdot \frac{\sqrt{4x+5}}{2\sqrt{x+1}} \, dx \]Cancel terms:\[ S = \pi \int_1^{11} \sqrt{4x+5} \, dx \]

Step 6: Evaluate the Integral

Integrate \( \sqrt{4x+5} \):Use substitution where \( u = 4x+5 \), hence \( du = 4dx \) or \( dx = \frac{1}{4}du \).Substitute back into the integral:\[ \int \sqrt{4x+5} \, dx = \int \frac{1}{4}\sqrt{u} \, du \]This gives:\[ \frac{1}{4} \cdot \frac{2}{3} u^{3/2} = \frac{1}{6}(4x+5)^{3/2} \]Evaluate this from 1 to 11:\[ \left[ \frac{1}{6}(4x+5)^{3/2} \right]_1^{11} = \frac{1}{6}(49^{3/2} - 9^{3/2}) = \frac{1}{6}(343 - 27) = \frac{1}{6} \cdot 316 = \frac{158}{3} \]

Step 7: Calculate the Surface Area

Multiply by \( \pi \) to get the total surface area:\[ S = \pi \cdot \frac{158}{3} = \frac{158\pi}{3} \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composition of functions. It is particularly useful when dealing with complex functions or when performing calculus operations like finding derivatives. Imagine you have a function within a function, represented as \( g(x) = h(f(x)) \). The chain rule tells us that the derivative of \( g \) with respect to \( x \) is the derivative of \( h \) with respect to \( f(x) \) multiplied by the derivative of \( f \) with respect to \( x \).This can be written as: \( g'(x) = h'(f(x)) \cdot f'(x) \). In our problem, we have \( f(x) = (x+1)^{1/2} \). When applying the chain rule:

• We differentiate the outer function, which is \( (x+1)^{1/2} \).
• The derivative of \( (x+1)^{1/2} \) is \( \frac{1}{2}(x+1)^{-1/2} \).

This gives us \( f'(x) = \frac{1}{2\sqrt{x+1}} \). Notice how we break down the operation to get the derivative of the inner function and multiply it by the derivative of the outer function.

Definite Integral

A definite integral calculates the area under a curve from one point to another on the \( x \)-axis. It is represented by the integral sign with limits. The lower limit is the starting point, and the upper limit is the ending point. In our exercise, we are revolving the graph of the function about the \( x \)-axis, so we need to compute a definite integral to find the total surface area.Definite integrals have applications beyond finding areas. They are also used in solutions concerning accumulation quantities and finding averages. When working with integrals:

• Set limits: Define your interval, in this example \([1, 11]\).
• Integrand: The function inside the integral; here it is \( \sqrt{4x+5} \).

The result of a definite integral provides the net area between the curve and the x-axis, taking into account any regions above and below the axis within the interval.

Substitution Method

The substitution method, also known as u-substitution, is a powerful tool for simplifying integrals. It involves changing the integral into a simpler form by substituting part of the integrand with a single variable. This simplification is often crucial for evaluating more complex integrals.Here’s how substitution was used in our problem:

• We set \( u = 4x + 5 \), which means the derivative \( du = 4dx \).
• This rearranges to \( dx = \frac{1}{4} du \).
• We replaced \( \sqrt{4x+5} \) with \( \sqrt{u} \) in the integral.

By transforming the limits of integration as well, integration becomes more straightforward and we end up with a function in terms of \( u \). After evaluation, this substitution method lets us easily integrate to get our result.

Surface of Revolution Formula

To find the surface area generated when a curve is rotated around an axis, we use the surface of revolution formula. This involves calculus techniques to integrate along the curve and find the total surface area. For a function \( y = f(x) \), the formula when revolving around the \( x \)-axis is: \[ S = 2\pi \int_a^b f(x) \sqrt{1 + (f'(x))^2} \, dx \]In our example:

• Function: \( f(x) = (x+1)^{1/2} \)
• Interval: \([1, 11]\)
• Derivative: \( f'(x) = \frac{1}{2\sqrt{x+1}} \)

The formula accommodates the curve and accounts for its slope defined by the derivative, \( f'(x) \). Here, the term \( \sqrt{1 + (f'(x))^2} \) corrects for any 'tilt' of the curve, adjusting how the surface area is calculated as the curve rotates. Combining all these, we use integration to find the precise surface area generated.