Question

Let \(R\) be the region bounded by the following curves. Use the disk method to find the volume of the solid generated when \(R\) is revolved about the \(x\) -axis. \(y=\sin x, y=0,\) for \(0 \leq x \leq \pi\) (Recall that \(\sin ^{2} x=\frac{1}{2}(1-\cos 2 x)\)

Short answer

- The function is \(y=\sin x\). 2. What is the interval we are considering for the revolution? - The interval is \(0\leq x\leq\pi\). 3. What method are we using to find the volume of the solid? - We are using the disk method. 4. What is the trigonometric identity provided for simplifying the squared function? - The trigonometric identity is \(\sin^2{x}=\frac{1}{2}(1-\cos{2x})\). 5. What is the volume of the solid generated when the region \(R\) is revolved around the \(x\)-axis? - The volume of the solid is \(\frac{3\pi^2}{8}\).

Step-by-step solution

Identify the function, the interval, and the axis of revolution

We have the function \(y=\sin x\), the interval \(0\leq x\leq\pi\), and we are revolving around the \(x\)-axis.

Square the function and apply the given identity

We need to find the area of each disk by squaring the function and applying the identity: \((\sin x)^2 = \frac{1}{2}(1-\cos{2x})\).

Set up the integral

We can now set up the integral to find the total volume using the disk method. We have: \(V=\pi\int_{0}^{\pi}\left[\frac{1}{2}(1-\cos{2x})\right]^2 dx=\pi\int_{0}^{\pi}\left(\frac{1}{4}(1-\cos{2x})^2\right)dx\)

Expand the integrand

Let's expand the integrand to make it easier to integrate: \(\frac{1}{4}(1 - 2\cos{2x} + \cos^2{2x})\)

Use the double-angle identity for \(\cos^2{x}\)

To simplify the \(\cos^2{2x}\) term, we'll use the double-angle identity for \(\cos^2{x}\), which is \(\cos^2{x} = \frac{1}{2}\left(1+\cos{2x}\right)\). Therefore, our integrand becomes: \(\frac{1}{4}(1 - 2\cos{2x} + \frac{1}{2}(1+\cos{4x}))\) Which simplifies to: \(\frac{1}{8}(3 - 4\cos{2x} + \cos{4x})\)

Integrate the simplified expression

Now we can integrate the simplified expression: \(V = \pi\int_{0}^{\pi}\frac{1}{8}(3 - 4\cos{2x} + \cos{4x}) dx\) This integral can be split into three separate integrals: \(V = \frac{\pi}{8}\left[\int_{0}^{\pi} 3dx - 4\int_{0}^{\pi} \cos{2x}dx +\int_{0}^{\pi} \cos{4x} dx \right]\)

Evaluate the integrals

Let's evaluate each of the three integrals: \(\int_{0}^{\pi} 3dx =[3x]_{0}^{\pi} = 3\pi - 3\cdot{0} = 3\pi\) \(\int_{0}^{\pi} \cos{2x}dx = [\frac{1}{2}\sin{2x}]_{0}^{\pi}=\frac{1}{2}[\sin{2\pi} - \sin{0}]=0\) \(\int_{0}^{\pi} \cos{4x} dx = [\frac{1}{4}\sin{4x}]_{0}^{\pi} =\frac{1}{4}[\sin{4\pi} - \sin{0}]=0\) Therefore, \(V = \frac{\pi}{8}\left[3\pi \right]\)

Final answer

Finally, we get the volume of the solid generated when region \(R\) is revolved around the \(x\)-axis: \(V = \frac{3\pi^2}{8}\)

Key concepts

Volume of Solids of Revolution

The volume of solids of revolution deals with creating a 3D shape by rotating a 2D area around an axis. When we say "disk method," we're talking about slicing this 3D shape into thin, flat disks. The idea is simple: if you have a curve, like the graph of a function, you can spin it around an axis—typically the x or y-axis—to generate a volume.
To calculate the volume using the disk method, you:

• Identify the region you want to rotate, defined by a function and intervals.
• Think of the region as being made up of numerous disks perpendicular to the axis of rotation.
• Squaring the function gives us the radius of each disk.

Finally, the formula for volume involves integrating the radii squared, multiplied by \(\pi\), taken over the interval. For example, in our case, revolving the curve \(y = \sin x\) around the x-axis results in the volume represented by the integral of these squared disks, summed over the interval from \(x = 0\) to \(x = \pi\). This turns the region between the curve and the x-axis into a solid, whose volume we calculated.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are essential in many areas of mathematics, especially when dealing with periodic phenomena. These functions are based on a circle, which relates nicely to the disk method as we're often revolving parts of a circle or its elements around an axis.
The function \(y = \sin x\) defines a wave-like pattern that oscillates between 1 and -1. In the interval from 0 to \(\pi\), \(\sin x\) smoothly moves from 0, peaks at 1 in the middle, and returns to 0. This symmetry is useful in both calculating an area under a curve and simplifying integrals.
An important identity we used is: \((\sin x)^2 = \frac{1}{2}(1 - \cos 2x)\). This identity emerged from a double-angle formula, which is crucial for converting squares of trigonometric functions into something easier to integrate. This conversion reduces complexity, making it more straightforward to plug into our methods like the disk method or other calculus approaches.

Integration Techniques

Integration is a powerful calculus tool used to find areas, volumes, and other important quantities. It involves summing up infinitesimally small quantities to obtain a total value, as in the case of finding the volume of a solid of revolution.
One common technique is using identities to simplify expressions before integrating. For instance, the identity \((\sin x)^2 = \frac{1}{2}(1 - \cos 2x)\) allowed us to transform a square into a simpler polynomial form, such as \((1 - \cos 2x)^2\).
Additionally, we split the integral, \(\pi\int_{0}^{\pi}\frac{1}{8}(3 - 4\cos{2x} + \cos{4x}) dx\), into separate, easier-to-handle integrals. Each of these can be evaluated independently before recombining to find the entire volume. Two key steps in integration:

• Recognizing and applying trigonometric identities to transform and simplify the integrand.
• Breaking down complex integrals into sums or differences of simpler integrals.

Mastering these integration techniques can make approaching complex calculus problems easier and more intuitive.