Problem 219 - Calculus Volume 1

Question

Write an integral that quantifies the change in the area of the surface of a cube when its side length doubles from s unit to 2 s units and evaluate the integral.

Step-by-Step Solution

Answer

The change in surface area is quantified by the integral as \( 18s^2 \).

Step 1: Determine Surface Area Formulas

The surface area of a cube with side length \( s \) is given by the formula \( A(s) = 6s^2 \). Similarly, for a cube with side length \( 2s \), the surface area becomes \( A(2s) = 6(2s)^2 = 24s^2 \).

Step 2: Define Change in Surface Area

The change in surface area as the side length changes from \( s \) to \( 2s \) is \( A(2s) - A(s) = 24s^2 - 6s^2 = 18s^2 \).

Step 3: Set Up the Integral

To understand this change via integration, consider the surface area as a function of \( x \), where \( x \) varies from \( s \) to \( 2s \). We use the integral \( \int_{s}^{2s} \frac{d}{dx}(6x^2) \, dx \) to represent the change.

Step 4: Calculate the Derivative

The derivative \( \frac{d}{dx}(6x^2) = 12x \).

Step 5: Integrate the Derivative Over the Interval

Compute the integral \( \int_{s}^{2s} 12x \, dx \). Evaluate by calculating \( \left[ 6x^2 \right]_{s}^{2s} \).

Step 6: Evaluate the Definite Integral

Substitute the bounds into the antiderivative: \( 6(2s)^2 - 6(s)^2 = 24s^2 - 6s^2 = 18s^2 \).

Step 7: Verify and Conclude

The calculated integral \( 18s^2 \) matches the previously calculated change in surface area, confirming our solution is correct.

Key Concepts

Surface Area

Surface area measures the total area of all the surfaces of a three-dimensional object. In the context of a cube, this is fairly straightforward. A cube has six identical square faces.
To find the total surface area of a cube with side length \( s \), you calculate the area of one square face, which is \( s^2 \), and then multiply by six (the number of faces). Thus, the surface area formula for a cube is \( A(s) = 6s^2 \).
When you increase the side length of a cube, the surface area changes. Specifically, if you double the side length from \( s \) to \( 2s \), the new area becomes \( A(2s) = 24s^2 \). This illustrates how the surface area grows dramatically with changes in the side length.

Definite Integral

A definite integral represents the net area under a curve between two specified bounds on a graph. It's a fundamental concept in calculus, allowing you to calculate total change or accumulation over an interval.
For the problem at hand, we use a definite integral to calculate the change in the surface area of the cube as its side length transitions from \( s \) to \( 2s \).
This involves integrating a derivative over this specific range, effectively giving us the net change in surface area, which is a perfect example of how integration quantifies change.

Cube Geometry

Understanding cube geometry is essential here. A cube is a special type of rectangular prism where all sides are equal. Therefore, it has some unique properties:

• Six identical square faces
• All angles are right angles
• Equal edges, so if one edge changes, all do

This uniformity simplifies calculations of geometric properties like surface area and volume. In this exercise, the primary focus is the surface area, which increases significantly as the side length increases. Doubling the side length quadruples the surface area, illustrating the non-linear growth characteristic of geometric figures.

Derivative Calculation

Calculating derivatives involves finding how a function changes as its input changes. Here, we derive the function \( A(x) = 6x^2 \), which represents the surface area relative to the side length \( x \).
The derivative, \( \frac{d}{dx}(6x^2) = 12x \), tells us how quickly the surface area increases as \( x \) increases. Derivatives are a core component of calculus and are crucial in understanding rates of change.
In the context of our problem, this derivative is integrated over an interval to compute the total change in surface area as we modify the cube's side length, showcasing the seamless transition from differential to integral calculus.