Problem 31
Question
Let \(R\) be the region bounded by the \(x\) axis, the curve \(y=\cos x+\sin x\), and the lines \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{6}\). Find the volume of the solid generated by rotating \(R\) about the \(x\) axis.
Step-by-Step Solution
Verified Answer
Volume is evaluated using the disk method over the given interval.
1Step 1 - Set Up the Integral
To find the volume of the solid generated by rotating the region around the x-axis, use the disk method. The formula for the volume is
2Step 2 - Identify the Limits of Integration
The limits of integration are given by the bounds in the problem. Here, the region is bounded between
3Step 3 - Define the Radius of Revolution
The radius of revolution for the disk method is given by the distance from the x-axis to the curve. In this problem, the radius is the function value of the curve, which is
4Step 4 - Formulate the Integral
The volume can be found by integrating the area of each disk along the interval, which results in the integral
5Step 5 - Evaluate the Integral
Evaluate the definite integral to find the volume. Simplify the integral and solve it by performing step-by-step integration and simplification.
Key Concepts
Disk MethodDefinite IntegralBounds of IntegrationRadius of RevolutionCurve Rotation
Disk Method
The Disk Method is a technique to find the volume of a solid formed by rotating a region around an axis. Imagine slicing the solid into thin disks perpendicular to the axis of rotation. Each disk has a small thickness and a circular face. The volume of each disk can be approximated as the volume of a cylinder.
The volume of a disk is given by \(\text{Volume} = \pi \times (\text{radius})^2 \times (\text{thickness})\). To find the total volume, we sum up the volumes of all disks, which leads us to an integral.
This method works particularly well when rotating a function around the x-axis or y-axis. When we rotate around the x-axis, the radius is the function value at a given point.
The volume of a disk is given by \(\text{Volume} = \pi \times (\text{radius})^2 \times (\text{thickness})\). To find the total volume, we sum up the volumes of all disks, which leads us to an integral.
This method works particularly well when rotating a function around the x-axis or y-axis. When we rotate around the x-axis, the radius is the function value at a given point.
Definite Integral
A definite integral is used to precisely calculate the area under a curve between two points. In the context of the Disk Method, a definite integral helps us find the total volume by summing up the infinite number of infinitesimally thin disks.
The notation for a definite integral is \[ \int_{a}^{b} f(x) \, dx \]. Here, \(a\) and \(b\) are the bounds of integration (limits between which we sum), and \( f(x) \) is the function we're integrating.
In our example, we set up the integral for the volume of the solid using the formula \(\text{Volume} = \int_{a}^{b} \pi [f(x)]^2 \, dx\). This integral provides the summed volume of all disks from \(x=-\frac{\pi}{2}\) to \(x=\frac{\pi}{6}\).
The notation for a definite integral is \[ \int_{a}^{b} f(x) \, dx \]. Here, \(a\) and \(b\) are the bounds of integration (limits between which we sum), and \( f(x) \) is the function we're integrating.
In our example, we set up the integral for the volume of the solid using the formula \(\text{Volume} = \int_{a}^{b} \pi [f(x)]^2 \, dx\). This integral provides the summed volume of all disks from \(x=-\frac{\pi}{2}\) to \(x=\frac{\pi}{6}\).
Bounds of Integration
The bounds of integration are the limits between which we calculate the definite integral. They define the interval over which we sum the volumes of the disks for rotation.
In the problem, the region is bounded by the lines \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{6}\). These values serve as our lower and upper bounds of integration.
Thus, for the definite integral, we integrate from \(x=-\frac{\pi}{2}\) to \(x=\frac{\pi}{6}\). Ensuring the correct bounds is crucial, as it dictates the span over which the function is summed.
In the problem, the region is bounded by the lines \(x=-\frac{\pi}{2}\) and \(x=\frac{\pi}{6}\). These values serve as our lower and upper bounds of integration.
Thus, for the definite integral, we integrate from \(x=-\frac{\pi}{2}\) to \(x=\frac{\pi}{6}\). Ensuring the correct bounds is crucial, as it dictates the span over which the function is summed.
Radius of Revolution
The radius of revolution refers to the distance from the axis of rotation to the curve we're rotating. This radius forms the circular cross-sections of the disks.
For the Disk Method, the radius at any point \(x\) is determined by the value of the function at \(x\). In this problem, the curve is \(y = \cos x + \sin x\), so the radius of a disk at any \(x\) is given by \(\cos x + \sin x\).
The radius plays a key role in determining the volume of each disk, as the area of the disk's face is proportional to the square of the radius.
For the Disk Method, the radius at any point \(x\) is determined by the value of the function at \(x\). In this problem, the curve is \(y = \cos x + \sin x\), so the radius of a disk at any \(x\) is given by \(\cos x + \sin x\).
The radius plays a key role in determining the volume of each disk, as the area of the disk's face is proportional to the square of the radius.
Curve Rotation
Curve rotation involves revolving a given region around an axis to form a three-dimensional solid. In our exercise, we rotate the function \(y=\cos x + \sin x\) around the x-axis.
This rotational movement creates a solid with circular cross-sections perpendicular to the x-axis. The Disk Method helps calculate the volume of this solid by summing the volumes of all disks along the interval.
In the integral setup, \(\pi [f(x)]^2 \) represents the area of each disk, and integrating this expression sums up the volumes from \(x = -\frac{\pi}{2}\) to \(x = \frac{\pi}{6}\).
This rotational movement creates a solid with circular cross-sections perpendicular to the x-axis. The Disk Method helps calculate the volume of this solid by summing the volumes of all disks along the interval.
In the integral setup, \(\pi [f(x)]^2 \) represents the area of each disk, and integrating this expression sums up the volumes from \(x = -\frac{\pi}{2}\) to \(x = \frac{\pi}{6}\).
Other exercises in this chapter
Problem 29
Use your calculator to solve the equation \(2 \tan 3 x-5.87=2 \sin 2 x \quad\) for \(0 \leq x \leq \frac{\pi}{2}\) to three decimal places.
View solution Problem 30
Find the area of the region bounded by the curves \(y=\sin 2 x\) and \(y=\cos x\) over the interval \(\frac{\pi}{6} \leq x \leq \frac{\pi}{2}\)
View solution Problem 32
a. Find the period \(p\), the amplitude \(b\), the horizontal shift \(d\), and the vertical shift \(a\) of the function \(f(x)=5.0+3.0 \cos \left[\frac{\pi}{4}(
View solution Problem 33
a. Find the period \(p\), the amplitude \(b\), the horizontal shift \(d\), and the vertical shift \(a\) of the function \(f(x)=33+27 \cos \left[\frac{2 \pi}{25}
View solution