Problem 47 - Calculus Early Transcendentals

Question

Find the volume of the solid of revolution. Sketch the region in question. The region bounded by \(y=1 / \sqrt{x}, y=0, x=2,\) and \(x=6\) revolved about the \(x\) -axis

Step-by-Step Solution

Answer

Question: Find the volume of the solid formed when the region bounded by y=1/√x, y=0, x=2, and x=6 is revolved about the x-axis. Answer: The volume of the solid of revolution is 𝜋 ln 3 cubic units.

Step 1: Sketch the region

Sketch the region bounded by \(y=1 / \sqrt{x}, y=0, x=2,\) and \(x=6\) on the Cartesian plane. You should see that the region is a section of the area under the curve \(y=1 / \sqrt{x}\) between the vertical lines at \(x=2\) and \(x=6\).

Step 2: Set up the disk method integral

In the disk method, we take cross-sectional disks perpendicular to the axis of revolution, which in this case is the \(x\)-axis. The radius of each disk is equal to the value of \(y\) for the given \(x\). Since we are revolving around the \(x\)-axis, the radius will be the function \(y=1 / \sqrt{x}\). The area of a disk can be expressed as \(A(x) = \pi \cdot [\frac{1}{\sqrt{x}}]^2\), which simplifies to \(A(x) = \frac{\pi}{x}\). Now, we will set up the integral from \(x=2\) to \(x=6\) of the area function to find the volume: $$V=\int_{2}^{6} A(x) \, dx = \int_{2}^{6} \frac{\pi}{x} \, dx$$

Step 3: Evaluate the integral

To evaluate the integral, we first find the antiderivative of the integrand. The antiderivative of \(\frac{\pi}{x}\) is: $$\pi \int \frac{1}{x} \, dx = \pi (\ln|x| + C)$$ Now, we will apply the Fundamental Theorem of Calculus to evaluate the integral: $$V = \pi(\ln|6| - \ln|2|) = \pi \ln\frac{6}{2}=\pi \ln 3$$

Step 4: State the final answer

The volume of the solid of revolution for the given region is \(\pi \ln 3\) cubic units.

Key Concepts

Solid of Revolution

A solid of revolution is created when a two-dimensional area is rotated around a line, known as the axis of rotation. In this case, imagine a flat shape, like a region on a graph, being spun around an axis to produce a three-dimensional object. This method is often used in calculus to find the volume of objects created in this way. To understand how it works, envision the curve of a function revolving around an axis. The space it sweeps out is the solid of revolution. Key points to remember include:

• The axis of rotation (like the x-axis in this problem).
• The region being revolved (bounded in this example by various lines and curves).

Every solid of revolution will have a distinctive volumetric shape dependant on the original region and the rotation.

Disk Method

The Disk Method is a powerful technique for finding the volume of a solid of revolution. Imagine slicing the solid into thin disks, akin to thin paper plates, and then stacking these disks to form the original volume.

• The axis of rotation determines the orientation of the disks.
• Each disk has a radius equal to the function value at a given point, forming a circle when revolved.

In the exercise, we defined the radius of each disk as the function value, y, from the curve being \(y=1/\sqrt{x}\). Thus, the area of a disk is \(\pi [1/\sqrt{x}]^2 = \frac{\pi}{x}\). Stack these disks from \(x=2\) to \(x=6\) using integration to find the volume.

Definite Integral

The definite integral plays a crucial role in calculus, especially when calculating the volume using the Disk Method. An integral calculates the accumulation of quantities, which, in this context, corresponds to summing the volumes of infinitesimally thin disks.To set up a definite integral, you need:

• A function that represents the radius or area of each disk, such as \(rac{\pi}{x}\).
• Bounds for the integration, specifically from where to where the disks stretch along the axis, noted here as \(x=2\) to \(x=6\).

The integral \(\int_{2}^{6} \frac{\pi}{x} \, dx\) accumulates the total volume from start to finish of the given range.

Antiderivatives

Antiderivatives, or indefinite integrals, are fundamental for solving definite integrals as they help determine the area underneath curves. An antiderivative of a function is a function whose derivative gives the original function.The process of finding antiderivatives is crucial for calculating integrals. In this exercise, the antiderivative of \(\frac{1}{x}\) becomes \(\ln|x| + C\), where \(C\) is a constant of integration.Once found, the Fundamental Theorem of Calculus allows you to evaluate the definite integral by assessing the difference of the antiderivative evaluated at the upper and lower bounds. This captures the total accumulation, giving the volume of the solid of revolution as \(\pi \ln 3\) in this instance.