Problem 222 - Calculus Volume 1

Question

Write an integral that quantifies the increase in the volume of a sphere as its radius doubles from \(R\) unit to 2\(R\) units and evaluate the integral.

Step-by-Step Solution

Answer

The increase in volume is \( \frac{28}{3} \pi R^3 \).

Step 1: Understanding the Problem

To find the increase in the volume of a sphere as its radius doubles, we need to evaluate the volume of a sphere of radius 2R and subtract the volume of a sphere of radius R. The formula for the volume of a sphere with radius r is \( V(r) = \frac{4}{3} \pi r^3 \).

Step 2: Setting Up the Integral

The increase in volume is represented by the integral of the derivative of the volume function with respect to the radius, evaluated from \(R\) to \(2R\). This can be written as: \[ \int_{R}^{2R} \frac{dV}{dr} \, dr \]

Step 3: Calculating the Derivative

Calculate the derivative of the volume function \( V(r) = \frac{4}{3} \pi r^3 \). The derivative \( \frac{dV}{dr} \) is \( 4 \pi r^2 \).

Step 4: Evaluating the Integral

Substitute the derivative into the integral: \[ \int_{R}^{2R} 4 \pi r^2 \, dr \]. Evaluate the integral: \[ 4 \pi \int_{R}^{2R} r^2 \, dr \].

Step 5: Finding the Antiderivative

Find the antiderivative of \( r^2 \), which is \( \frac{r^3}{3} \). So the integral becomes \( 4 \pi [ \frac{r^3}{3} ] \).

Step 6: Substituting Limits

Substitute the limits of integration \(R\) and \(2R\) into the antiderivative: \[ 4 \pi \left( \frac{(2R)^3}{3} - \frac{R^3}{3} \right) \].

Step 7: Simplifying the Result

Simplify the expression: \[ 4 \pi \left( \frac{8R^3}{3} - \frac{R^3}{3} \right) = 4 \pi \left( \frac{7R^3}{3} \right) = \frac{28}{3} \pi R^3 \].

Key Concepts

Volume of a Sphere

The concept of finding the volume of a sphere is central to many problems in calculus, including our example where we determine the change in volume as the radius doubles. The formula for the volume of a sphere is derived using integration and is given by: \[ V(r) = \frac{4}{3} \pi r^3 \]Here, \( r \) represents the radius of the sphere. This formula allows us to calculate the space inside a sphere, an essential step if you want to understand how the volume changes with the size of the sphere.
In our exercise, we used this formula to determine the initial and final volumes when the radius is \( R \) and \( 2R \), respectively. By subtracting the volume of the sphere when the radius is \( R \) from the volume at \( 2R \), we can find the increase in volume. This technique highlights the practical application of the volume formula in solving real-world problems.

Antiderivative

Antiderivatives are central to integral calculus and are often used to find areas under curves, as well as volumes under surfaces. In essence, an antiderivative of a function \( f(x) \) is a function \( F(x) \) whose derivative is \( f(x) \).
In our exercise, we aimed to find the increase in volume of a sphere by integrating. During this process, we encountered the function \( 4 \pi r^2 \) as the derivative of the volume function \( V(r) \).
The antiderivative of this function, \( r^2 \), is \( \frac{r^3}{3} \). By finding this antiderivative, we transformed the expression \( 4 \pi \int_{R}^{2R} r^2 \, dr \) into a simpler one that we could easily evaluate with the given limits. Understanding antiderivatives is fundamental for solving definite integrals, like in our problem.

Definite Integral

A definite integral is a powerful tool in calculus that helps find the accumulation of quantities, such as areas or volumes. In our problem, it measured the increase in the volume of a sphere when the radius changes from \( R \) to \( 2R \). The method involves setting up an integral with specific limits; in our case, from \( R \) to \( 2R \). This looks like:\[ \int_{R}^{2R} \frac{dV}{dr} \, dr \]After substituting the derivative \( 4 \pi r^2 \) in place of \( \frac{dV}{dr} \), the integral becomes \( 4 \pi \int_{R}^{2R} r^2 \, dr \).
To solve it, we found the antiderivative, calculated it at the limits, and subtracted. Substituting our limits yielded:\[ 4 \pi \left( \frac{(2R)^3}{3} - \frac{R^3}{3} \right) \]This definite integral gave us a precise measure of how much the sphere's volume increased as the radius doubled, resulting in \( \frac{28}{3} \pi R^3 \). Mastery of definite integrals is crucial for solving many physical problems where changes between two states are measured.