Problem 26 - Single Variable Calculus

Question

In Exercises \(13-34\), find the volume of the solid generated by revolving the region bounded by the graphs of the equations and/or inequalities about the indicated axis. $$ y=\sqrt{\sin x}, \quad y=0, \quad x \leq \frac{\pi}{2} ; \quad \text { the } x \text { -axis } $$

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region bounded by \(y=\sqrt{\sin{x}}\), \(y=0\), and \(x\leq\frac{\pi}{2}\) around the \(x\)-axis is \(V=\pi\int_{0}^{\frac{\pi}{2}}\sin{x}\, dx = \pi\) cubic units.

Step 1: Identify the region and the method to use

Sketch the graphs of the given curves, and find the points at which the graph of \(y=\sqrt{\sin{x}}\) intersects with the \(x\)-axis. Since the region is bounded by the \(x\)-axis and the curve \(y = \sqrt{\sin x}\), we can use the disk method for finding the volume.

Step 2: Set up the integral for the volume using the disk method

The formula for the volume V using the disk method is given by: \[V = \pi\int_{a}^{b}[R(x)]^2 dx\] where \(a\) and \(b\) are the limits of integration and \(R(x)\) is the distance from the curve to the axis of revolution (in this case, the \(x\)-axis). Here, \(R(x) = \sqrt{\sin x}\), because the curve is already expressed in the form \(y=f(x)\).

Step 3: Determine the integration limits

According to the given conditions, \(x\) should be smaller than or equal to \(\frac{\pi}{2}\), so the limits of integration are \(0 \leq x \leq \frac{\pi}{2}\). Thus, our integral becomes: \[V = \pi\int_{0}^{\frac{\pi}{2}}(\sqrt{\sin x})^2 dx\]

Step 4: Evaluate the integral

Simplify the integral: \[V = \pi\int_{0}^{\frac{\pi}{2}}\sin x \, dx\] \[V = \pi[-\cos x\big\rvert_{0}^{\frac{\pi}{2}}]\] \[V = \pi[-\cos\left(\frac{\pi}{2}\right) + \cos(0)] = \pi[0+1] = \pi\] The volume of the solid generated by revolving the region bounded by the given curves around the \(x\)-axis is \(\pi\) cubic units.

Key Concepts

Disk method

The Disk Method is a technique used in calculus to find the volume of a solid of revolution. This method is particularly useful when a region in the plane is revolved around a line (usually an axis). The result is a 3D shape whose volume can be determined through integration.
To use the disk method, imagine slicing the solid into thin disks perpendicular to the axis of revolution. The volume of each thin disk can be approximated by its area multiplied by its thickness.

• The area of a disk is given by \(\pi [R(x)]^2\), where \R(x)\ is the radius of the disk.
• The volume \(V\) of the entire solid can be found by summing up the volumes of these disks, which translates into solving an integral: \[V = \pi \int_{a}^{b} [R(x)]^2 \, dx\]

Using the disk method, we were able to calculate the volume of a solid generated by rotating a region bounded by \y = \sqrt{\sin x}\ about the \x\-axis.

Integration limits

Integration limits define the bounds of the region being considered in the problem. They are crucial because they determine the portion of the graph that is being revolved to form the solid of revolution.
The given integration limits need to encapsulate the entire area that is being considered for rotation without missing any part of the region.
In our problem, it's stated that \x \, \leq \, \frac{\pi}{2}\, which means we are only interested in the values of \x\ from \0\ to \frac{\pi}{2}\.

Using these limits, we can ensure that the integral captures the entire curve of \y = \sqrt{\sin x}\ from its starting point on the \x\-axis to its endpoint at \x = \frac{\pi}{2}\. Therefore, the limits of integration used are \[0 \, \leq \, x \, \leq \, \frac{\pi}{2}\].
These correct limits ensure the calculation of the entire intended volume.

Definite integral

A Definite Integral is a fundamental concept in calculus used to calculate the net area under a curve, and in this case, for determining the volume of a solid generated by a function.
With the definite integral, the process is straightforward:

• It compresses a sum of infinite tiny products (areas of disks) into a single value representing total volume.
• From the generic disk method integral setup: \[V = \pi \int_{a}^{b} [R(x)]^2 \, dx\]
• This setup becomes: \[V = \pi \int_{0}^{\frac{\pi}{2}} [\sqrt{\sin x}]^2 \, dx = \pi \int_{0}^{\frac{\pi}{2}} \sin x \, dx\]

This definite integral results in an exact volume of \pi\ cubic units after evaluating it through standard calculus techniques.
The definite integral simplifies the process of summing infinitesimals into useful quantities like volume, making it indispensable for such applications.