Problem 1 - Calculus Single Variable

Question

In each of Exercises 1-6, use the method of disks to calculate the volume \(V\) of the solid that is obtained by rotating the given planar region \(\mathcal{R}\) about the \(x\) -axis. \(\mathcal{R}\) is the region below the graph of \(y=\sqrt{x},\) above the \(x\) -axis, and between \(x=1\) and \(x=3\).

Step-by-Step Solution

Answer

The volume is \( 4\pi \).

Step 1: Understanding the Disk Method Formula

The volume of a solid of revolution generated by rotating a function around the x-axis is given by the integral formula \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \), where \(f(x)\) is the function representing the outer radius of the disks. In this problem, \(f(x) = \sqrt{x}\).

Step 2: Define the Boundaries of Integration

The region \( \mathcal{R} \) lies between \(x = 1\) and \(x = 3\), so the integral will be computed from 1 to 3.

Step 3: Set Up the Integral

Using the disk method formula, set up the integral: \[ V = \pi \int_{1}^{3} [\sqrt{x}]^2 \, dx \]. Since \([\sqrt{x}]^2 = x\), this simplifies to \[ V = \pi \int_{1}^{3} x \, dx \].

Step 4: Integrate the Function

Integrate \(x\) with respect to \(x\): \[ \int x \, dx = \frac{x^2}{2} \].

Step 5: Evaluate the Integral

Substitute the limits of integration into the antiderivative: \[ \pi \left[ \frac{x^2}{2} \right]_{1}^{3} = \pi \left( \frac{3^2}{2} - \frac{1^2}{2} \right) \].

Step 6: Calculate the Final Value

Calculate the numeric value: \( \pi \left( \frac{9}{2} - \frac{1}{2} \right) = \pi \times 4 = 4\pi \).

Step 7: Present the Final Answer

The volume \( V \) of the solid obtained by rotating the region \( \mathcal{R} \) about the x-axis is \( 4\pi \).

Key Concepts

Volume of Solid of Revolution

The volume of a solid of revolution is a fascinating concept in calculus that deals with finding the volume of three-dimensional objects formed by rotating a two-dimensional shape around an axis. In our case, we use the "Disk Method," which is particularly effective when rotating a region around the x-axis. Imagine slicing the shape into many thin, flat disks. Each disk has a small thickness, represented by the change in x, denoted as \(dx\). To find the volume, you sum up the volume of all these infinitesimally thin disks. The formula used for this purpose is:

• \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \)

The variable \(f(x)\) represents the function that describes the curve, which in our scenario is \(y = \sqrt{x}\). So, the disks have radii equal to \(\sqrt{x}\), making the area \(\pi [\sqrt{x}]^2 = \pi x\). In simple terms, you're stacking up all these disk volumes from the starting point \(x=1\) to the endpoint \(x=3\) to get the total volume.

Calculus Integration

Integration is one of the two fundamental operations in calculus, the other being differentiation. While differentiation looks at rates of change, integration helps us understand the accumulation of quantities. In the context of finding the volume using the disk method, integration helps us sum the infinite number of small volumes of the disks along the x-axis. When you look at the integral \[ \int_{1}^{3} x \, dx \], it symbolizes this accumulation. In essence, integration here is about compiling all the tiny disk volumes to form one complete volume.The process involves finding an antiderivative, which is a function that tells us how the area under the graph of \(y = x\) accumulates as \(x\) increases:

• \( \int x \, dx = \frac{x^2}{2} \)

This expression \(\frac{x^2}{2}\) gives us a sense of the size of these accumulated areas as the diameter changes from \(x = 1\) to \(x = 3\). In everyday terms, integration calculates how the tiny contributions from all parts of \(x\)'s range add to the whole.

Definite Integral Calculation

Definite integrals are a straightforward way to compute the total accumulation of a quantity over an interval. In this problem, we're dealing with a definite integral from \(x=1\) to \(x=3\). The boundaries of the integral, represented as \([a, b]\), tell us the start and stop points for our summation. The calculation goes beyond finding just an antiderivative. After calculating the antiderivative \(\frac{x^2}{2}\), we must evaluate its value at the upper bound (\(x=3\)) and subtract the value at the lower bound (\(x=1\)):

• \( \pi \left[ \frac{x^2}{2} \right]_{1}^{3} = \pi \left( \frac{9}{2} - \frac{1}{2} \right) \)

The difference describes how much area the disks encompass over the entire interval from 1 to 3. Performing this subtraction ensures we measure only the intended region, not any beyond our precise limits. Ultimately, this valuation gives us a numerical result, \(4\pi\), which represents the solid's total volume. In real-life applications, this approach allows for exact calculations of spaces under curves or within bounded regions, essential for everything from engineering to physics.