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Magazine Chapter 6: Problem 21
Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(x\)-axis. \(y=x^{2}, \quad y=0, \quad x=2\)
Short Answer
Expert verified
The volume is \(\frac{32\pi}{5}\).
Step by step solution
01
Understand the Boundaries
We are given the region bounded by the curve \(y = x^2\), the line \(y = 0\) (which is the x-axis), and the vertical line \(x = 2\). We need to visualize this region in the xy-plane before performing any calculations.
02
Apply Disc Method Formula
To find the volume of the solid formed by revolving the region around the x-axis, we use the disc method. The volume \(V\) is given by the formula: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] where \(f(x) = x^2\), and the limits \(a\) and \(b\) are 0 and 2, respectively.
03
Set Up the Integral
Plug the function \(f(x) = x^2\) and the limits \(0\) and \(2\) into the integral formula: \[ V = \pi \int_{0}^{2} (x^2)^2 \, dx \] which simplifies to \[ V = \pi \int_{0}^{2} x^4 \, dx \].
04
Evaluate the Integral
Find the integral of \(x^4\) from 0 to 2: \[ \int x^4 \, dx = \frac{x^5}{5} + C \]. Evaluate this from 0 to 2: \[ \left. \frac{x^5}{5} \right|_{0}^{2} = \frac{2^5}{5} - \frac{0^5}{5} = \frac{32}{5}. \]
05
Calculate the Volume
Multiply the result of the integral by \(\pi\) to find the volume: \[ V = \pi \times \frac{32}{5} = \frac{32\pi}{5}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Disc Method
The disc method is a popular technique in calculus for finding the volume of a solid of revolution. Imagine slicing the solid into thin discs perpendicular to the axis of rotation. Each of these discs has a circular face.
The volume of each disc can be expressed using the formula for the area of a circle, which is \( ext{Area} = \pi r^2 \), where \( r \) is the radius of the disc. For our problem, the radius \( r \) is defined by the function \( f(x) = x^2 \).
By adding up (or integrating) all these tiny disc volumes from \( x = a \) to \( x = b \), we get the complete volume of the solid. That's why the formula for the disc method involves an integral:
The volume of each disc can be expressed using the formula for the area of a circle, which is \( ext{Area} = \pi r^2 \), where \( r \) is the radius of the disc. For our problem, the radius \( r \) is defined by the function \( f(x) = x^2 \).
By adding up (or integrating) all these tiny disc volumes from \( x = a \) to \( x = b \), we get the complete volume of the solid. That's why the formula for the disc method involves an integral:
- The integral bounds \([a, b]\) define the segment along the axis being considered.
- The squared function \([f(x)]^2\) gives the shape of the disc's face.
- The \( \pi \) constant arises from the circular cross-section.
Revolving Regions
Revolving regions is a process where you rotate a defined region about an axis to create a three-dimensional object, or a solid of revolution.
In practice, this involves a 2D region, like between a curve and an axis, being spun around a line (typically the x-axis or y-axis) to fill a 3D space.
In this problem, we revolve the region bounded by \( y = x^2 \), \( y = 0 \), and \( x = 2 \) around the x-axis. This rotation transforms the flat area into a solid.
In practice, this involves a 2D region, like between a curve and an axis, being spun around a line (typically the x-axis or y-axis) to fill a 3D space.
In this problem, we revolve the region bounded by \( y = x^2 \), \( y = 0 \), and \( x = 2 \) around the x-axis. This rotation transforms the flat area into a solid.
- This is why the starting line \( y = 0 \) is significant: it acts as the axis of revolution.
- The line \( x = 2 \) defines the endpoint of the region we'll revolve, signifying the end of our volume calculations.
- Understanding these boundaries is crucial for accurately setting up the integral that calculates volume.
Definite Integral
A definite integral is a type of integral that provides the accumulated quantity, such as area under a curve, over a specific interval \([a, b]\).
For volume calculations involving rotation, definite integrals help compute the exact size of the 3D object formed.
In our specific problem, the definite integral \( \int_{0}^{2} x^4 \, dx \) measures the accumulated disc areas between \( x = 0 \) and \( x = 2 \).
For volume calculations involving rotation, definite integrals help compute the exact size of the 3D object formed.
In our specific problem, the definite integral \( \int_{0}^{2} x^4 \, dx \) measures the accumulated disc areas between \( x = 0 \) and \( x = 2 \).
- The function inside the integral, \( x^4 \), comes from squaring the original curve function \( f(x) = x^2 \).
- After integration, limits \( a \) and \( b \) are plugged into the integral's result to get the final answer.
- The final numerical result of this integral provides the base for calculating the total volume of the revolution solid.
Integration Techniques
Integration techniques involve methods to find the integral of a function, which is the inverse operation to differentiation.
In this exercise, we used a straightforward power-rule integration:
For a polynomial like \( x^4 \), the antiderivative is \( \frac{x^5}{5} \).
Let's see steps for using this technique:
In this exercise, we used a straightforward power-rule integration:
For a polynomial like \( x^4 \), the antiderivative is \( \frac{x^5}{5} \).
Let's see steps for using this technique:
- Increase the exponent of \( x \) by one.
- Divide by the new exponent to get the antiderivative.
- Don't forget to apply the limits of integration to compute the definite integral correctly.
