Problem 32 - Calculus Single Variable

Question

Calculate the area \(S\) of the surface obtained when the given function, over the given interval, is rotated about the \(y\) -axis. $$ f(x)=x^{2} \quad [0, \sqrt{2}] $$

Step-by-Step Solution

Answer

The surface area is \( \frac{26\pi}{3} \).

Step 1: Set up Formula for Surface Area

To calculate the surface area of a curve rotated about the y-axis, use the formula \( S = \int_a^b 2\pi x \sqrt{1 + \left(\frac{dy}{dx}\right)^2 } \, dx \). For the function \( f(x) = x^2 \), we have \( y = x^2 \).

Step 2: Compute Derivative

Calculate the derivative \( \frac{dy}{dx} \) for \( y = x^2 \), which gives \( \frac{dy}{dx} = 2x \).

Step 3: Substitute into Surface Area Formula

Substitute \( x^2 \) and \( \frac{dy}{dx} = 2x \) into the surface area formula: \( S = \int_0^{\sqrt{2}} 2\pi x \sqrt{1 + (2x)^2} \, dx \). This simplifies to \( S = \int_0^{\sqrt{2}} 2\pi x \sqrt{1 + 4x^2} \, dx \).

Step 4: Simplify and Evaluate the Integral

Simplify the integral \( S = 2\pi \int_0^{\sqrt{2}} x \sqrt{1 + 4x^2} \, dx \). Use substitution: let \( u = 1 + 4x^2 \), then \( du = 8x \, dx \), or \( x \, dx = \frac{1}{8} \, du \). Change the limits according to \( u \): when \( x = 0 \), \( u = 1 \), and when \( x = \sqrt{2} \), \( u = 9 \).

Step 5: Integrate with New Variables

Substitute \( x \, dx = \frac{1}{8} \, du \) and \( u \) into the integral: \( S = 2\pi \int_1^9 \frac{1}{8} u^{1/2} \, du \). This simplifies to \( S = \frac{\pi}{4} \int_1^9 u^{1/2} \, du \).

Step 6: Calculate the Indefinite Integral

Evaluate \( \frac{\pi}{4} \left[ \frac{2}{3} u^{3/2} \right]_1^9 \). Substitute the limits to get \( \frac{\pi}{4} \left[ \frac{2}{3}(9^{3/2}) - \frac{2}{3}(1^{3/2}) \right] \).

Step 7: Compute Final Answer

Calculate \( 9^{3/2} = 27 \) and \( 1^{3/2} = 1 \), so the expression becomes \( \frac{\pi}{6} [54 - 2] = \frac{\pi}{6} \times 52 = \frac{26\pi}{3} \).

Key Concepts

Rotating a Function

When a function or curve is rotated around an axis, it creates a three-dimensional shape. This process is known as solid of revolution. In this exercise, we are tasked with rotating a function around the y-axis. This specific rotation transforms two-dimensional graphs into three-dimensional objects.
For example, rotating the function \( y = x^2 \) from \([0, \sqrt{2}]\) around the y-axis, produces a paraboloid.
To visualize this, imagine taking the curve of \( y = x^2 \) and spinning it rapidly around the y-axis - it will generate a bowl-like surface. This concept is fundamental in applications where we want to calculate the surface area or volume of such solids.
The general formula used for calculating the surface area of a function rotated about the y-axis is:
\[ S = \int_a^b 2\pi x \sqrt{1 + \left(\frac{dy}{dx}\right)^2 } \, dx \]
This formula combines the width of the slice, represented by \(2\pi x\), with the length of the curve, represented through the square root term.

Integral Calculus

Integral calculus focuses on finding the total or accumulated change of a quantity, usually expressed as areas under curves. In the context of calculating surface areas of revolution, integration is essential. We used it to accumulate small pieces of surface area formed as we rotate the curve about the y-axis.
The key formula involved is an integral that sums an infinite number of infinitesimal surface areas defined by our rotated function.
When integrating, the limits \([0, \sqrt{2}]\) define the section of the curve we consider. Within these boundaries, every tiny piece along the x-axis contributes to the entire surface. This is expressed by:
\[ S = \int_0^{\sqrt{2}} 2\pi x \sqrt{1 + 4x^2} \, dx \]

• The integration turns into an evaluation problem by substituting values.
• In calculations, a substitution method is often helpful and is used here using \( u = 1 + 4x^2 \).

This allows us to simplify and solve the integral, bringing us a step closer to finding the exact surface area.

Derivative Calculation

A derivative represents the rate of change of a function with respect to one of its variables. For our surface area exercise, finding the derivative of the function \( y = x^2 \) was crucial.
The differentiation gives us \( \frac{dy}{dx} = 2x \), which tells us how steep the curve is at any point.
This derivative is plugged into the original surface area formula to adjust for the shape and steepness of the curve as it pivots around the y-axis:
\[ S = \int_a^b 2\pi x \sqrt{1 + (2x)^2} \, dx \]

• Knowing the derivative helps in shaping the adjustments needed to accurately calculate the surface area.
• Derivatives are a streamlined way of understanding changes locally, which when combined through integration, turns into global information.

In our task, after calculating the derivative, it was incorporated into the integral to reflect the true nature of the surface created after rotation.