Problem 25 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(x\) -axis. \(y=\sqrt{\cos ^{-1} x}\), in the first quadrant

Step-by-Step Solution

Answer

Question: Determine the volume of the solid generated when the region R enclosed by the curves \(y=\frac{1}{\sqrt{1+x^{2}}}\), \(y=0\), \(x=-1\), and \(x=1\) is revolved about the x-axis. Answer: The volume of the solid generated when the region R is revolved about the x-axis is \(V = \frac{\pi^3}{2}\).

Step 1: Sketch the graph and identify the region R

To find the region R, we need to sketch the graph of the given curves: \(y=\frac{1}{\sqrt{1+x^{2}}}\), \(y=0\), \(x=-1\), and \(x=1\). Then, we can see the enclosed area which is the region R we want to revolve around the x-axis.

Step 2: Use disk method to find the volume

To find the volume of the solid formed by revolving the region R about the x-axis, we will be using the disk method which is: \( V = \pi \int_a^b [(\text{outer radius})^2 - (\text{inner radius})^2] dx \) In this case, we have the outer radius as y and inner radius as 0. So the formula for this problem becomes: \( V = \pi \int_{-1}^1 [y^2] dx \) Now substitute y with the given equation: \( V = \pi \int_{-1}^1 [\left(\frac{1}{\sqrt{1+x^{2}}}\right)^2] dx \)

Step 3: Simplify the integration and setup the integral

Now, we need to simplify the integration before solving it: \( V = \pi \int_{-1}^1 [\frac{1}{1+x^2}] dx \)

Step 4: Solve the integral

Now, we can solve the integral. The integration of \(\frac{1}{1+x^2}\) is \(\arctan(x)\): \( V = \pi [\arctan(x)]_{-1}^1 \)

Step 5: Calculate the volume of the solid

To calculate the volume of the solid, we will evaluate the expression we obtained in the previous step with the limits: \( V = \pi [\arctan(1) - \arctan(-1)] \) We know \(\arctan(1) = \frac{\pi}{4}\) and \(\arctan(-1) = -\frac{\pi}{4}\): \( V = \pi [\frac{\pi}{4} - (-\frac{\pi}{4})] \) \( V = \pi [\frac{\pi}{2}] \) Finally, the volume of the solid generated is: \( V = \frac{\pi^3}{2} \)

Key Concepts

Solid of Revolution

A solid of revolution is a three-dimensional object created by rotating a two-dimensional plane area around a straight line (axis) that lies on the same plane. Visualizing this can sometimes be challenging, so here’s a helpful hint: imagine a piece of paper with a shape drawn on it. When you roll the paper around a pencil, the outline of that shape creates a three-dimensional form. That form is your solid of revolution.

In the given exercise, the region known as R is revolved around the x-axis. To generate a solid, we revolve a curve which in this case is bounded by the equation \(y=\frac{1}{\sqrt{1+x^{2}}}\), creating a symmetric solid with respect to the x-axis. It's essential to sketch this region first, which allows us to visually grasp the shape that will be rotated and hence the solid that will be formed.

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the total size or value, such as areas, volumes, and other quantities, when the size of a part is known. In essence, it deals with accumulation. The central concept in integral calculus is the definite integral, represented by the symbol \(\int\). When we set up an integral to find the volume of a solid of revolution, we are accumulating tiny slices or disks to form the entire solid.

The process often involves integrating functions which sometimes requires applying sophisticated techniques and understanding the function's behavior. Integral calculus is not only about setting up integrals correctly; it's also about being able to manipulate and evaluate them, which is a crucial step in volume calculation using the disk method.

Volume of Solid

When calculating the volume of a solid, especially one of revolution, we need to consider the solid as a series of thin, flat circular disks or washers. The disk method, which is applied in the exercise, stacks these infinitesimally thin disks along the axis of rotation. Each disk has a volume of \(\pi r^2\Delta x\), where \(r\) is the radius of the disk, and \(\Delta x\) is its width (thickness).

For the entire volume, we sum up the volumes of these individual disks from one end of the solid to the other, which translates to integrating the function that gives us the radius of the disks over the interval of interest, which, in this exercise, is from \(x=-1\) to \(x=1\).

Definite Integral Evaluation

The definite integral evaluation is the process of calculating the exact value of a definite integral. It represents the net signed area under the curve of a given function from one point to another. In the context of computing volumes, it gives us the accumulated measure, which, in our case, is the volume of the solid of revolution.

The specific evaluation carried out in the exercise involves finding the antiderivative of the function \(\frac{1}{1+x^2}\), which is \(\arctan(x)\), and then using the fundamental theorem of calculus, which states that the exact value of a definite integral is the difference of the values of its antiderivative evaluated at the upper and lower limits of integration. Hence by evaluating \(\arctan(x)\) at \(x=1\) and \(x=-1\), we determine the volume of the solid created by revolving our original function about the x-axis.