Problem 76
Question
Let \(R\) be the region bounded by the graph of \(f(x)=x^{-p}\) and the \(x\) -axis, for \(x \geq 1\) a. Let \(S\) be the solid generated when \(R\) is revolved about the \(x\) -axis. For what values of \(p\) is the volume of \(S\) finite? b. Let \(S\) be the solid generated when \(R\) is revolved about the \(y\) -axis. For what values of \(p\) is the volume of \(S\) finite?
Step-by-Step Solution
Verified Answer
Answer: The volume is finite when revolved about the x-axis for \(p > 1\) and when revolved about the y-axis for \(p > 3\).
1Step 1: Set up the integral for the volume using the disk method
To set up the integral, we use the disk method since the solid is generated by revolving the region R about the x-axis. The formula for the disk method is given by:
$$V = \pi\int_a^b [f(x)]^2 dx$$
For our function, \(f(x) = x^{-p}\), so \([f(x)]^2 = x^{-2p}\). The given boundaries are \(x \geq 1\), so the limits of integration are from 1 to \(\infty\).
The integral becomes:
$$V = \pi\int_1^{\infty} x^{-2p} dx$$
2Step 2: Evaluate the integral and check for convergence
To evaluate the integral, we use the rule for the integral of a power function:
$$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$
Applying this rule to our integral, we get:
$$V = \pi\left[\frac{x^{-2p+1}}{-2p+1}\right]_1^{\infty}$$
For the integral to converge and have a finite volume, the limit as \(x\) goes to infinity must be finite. Therefore, we need \(-2p+1 < -1\) or \(p > 1\).
In conclusion, the volume of the solid generated when R is revolved about the x-axis is finite for \(p > 1\).
#b. Revolving R about the y-axis#
3Step 1: Set up the integral for the volume using the shell method
To set up the integral, we use the shell method since the solid is generated by revolving the region R about the y-axis. The formula for the shell method is given by:
$$V = 2\pi\int_a^b x\cdot f(x) dx$$
For our function, \(f(x) = x^{-p}\), so \(x\cdot f(x) = x^{1-p}\). The given boundaries are \(x \geq 1\), so the limits of integration are from 1 to \(\infty\).
The integral becomes:
$$V = 2\pi\int_1^{\infty} x^{1-p} dx$$
4Step 2: Evaluate the integral and check for convergence
To evaluate the integral, we use the rule for the integral of a power function again:
$$\int x^n dx = \frac{x^{n+1}}{n+1} + C$$
Applying this rule to our integral, we get:
$$V = 2\pi\left[\frac{x^{2-p}}{2-p}\right]_1^{\infty}$$
For the integral to converge and have a finite volume, the limit as \(x\) goes to infinity must be finite. Therefore, we need \(2-p < -1\) or \(p > 3\).
In conclusion, the volume of the solid generated when R is revolved about the y-axis is finite for \(p > 3\).
Key Concepts
Disk MethodShell MethodImproper Integrals
Disk Method
The disk method is a powerful technique for finding volumes of solids of revolution. Imagine you have a simple 2D shape, and you spin it around an axis to form a 3D object. The disk method helps calculate the volume of such an object.
It's like stacking up a bunch of thin disks or cylinders to form the shape.
The formula for the disk method is:
This integral covers the range from \( a \) to \( b \).
Each slice, a tiny disk, has a volume of \( \pi [f(x)]^2 \). When you revolve a region around the x-axis, this process becomes quite handy.
For example, given the function \( f(x) = x^{-p} \) bounded by \( x \geq 1 \), if you revolve it about the x-axis, you get a solid. Using the disk method, you would find the volume by calculating the integral \( V = \pi \int_1^{\infty} x^{-2p} \, dx \).
This integral helps determine the conditions under which the volume is finite. If you look for conditions where the integral converges, it occurs when \( p > 1 \). So, for \( f(x) = x^{-p} \) spun around the x-axis, the solid has a finite volume for \( p > 1 \). This shows us how this method can predict how long and narrow shapes have finite or infinite volumes when revolved.
It's like stacking up a bunch of thin disks or cylinders to form the shape.
The formula for the disk method is:
- \( V = \pi \int_a^b [f(x)]^2 \, dx \)
This integral covers the range from \( a \) to \( b \).
Each slice, a tiny disk, has a volume of \( \pi [f(x)]^2 \). When you revolve a region around the x-axis, this process becomes quite handy.
For example, given the function \( f(x) = x^{-p} \) bounded by \( x \geq 1 \), if you revolve it about the x-axis, you get a solid. Using the disk method, you would find the volume by calculating the integral \( V = \pi \int_1^{\infty} x^{-2p} \, dx \).
This integral helps determine the conditions under which the volume is finite. If you look for conditions where the integral converges, it occurs when \( p > 1 \). So, for \( f(x) = x^{-p} \) spun around the x-axis, the solid has a finite volume for \( p > 1 \). This shows us how this method can predict how long and narrow shapes have finite or infinite volumes when revolved.
Shell Method
To explore volumes of solids of revolution from a different perspective, we use the shell method. Imagine wrapping a ribbon around your shape. Now, spin it around an axis to form a solid. This method comes handy when revolving around the y-axis.
Here, you picture the 2D shape as a series of cylindrical shells stacked side by side.
The formula used is:
This method is best when the axis of revolution is parallel to the height of the object, like the y-axis. Consider again the function \( f(x) = x^{-p} \).
When revolving around the y-axis, the shape creates a solid, with the volume calculated by \( V = 2\pi \int_1^{\infty} x^{1-p} \, dx \). The crucial part of this method is figuring out for which values of \( p \) the integral converges. Here, the volume is finite for \( p > 3 \).
This illustrates that the shell method handles scenarios where more intuitive methods like the disk method require complex transformations. The integration helps us evaluate which conditions will provide our solid a finite volume. Hence, the shell method becomes invaluable in varied problem scenarios of volumes of solids of revolution.
Here, you picture the 2D shape as a series of cylindrical shells stacked side by side.
The formula used is:
- \( V = 2\pi \int_a^b x \cdot f(x) \, dx \)
This method is best when the axis of revolution is parallel to the height of the object, like the y-axis. Consider again the function \( f(x) = x^{-p} \).
When revolving around the y-axis, the shape creates a solid, with the volume calculated by \( V = 2\pi \int_1^{\infty} x^{1-p} \, dx \). The crucial part of this method is figuring out for which values of \( p \) the integral converges. Here, the volume is finite for \( p > 3 \).
This illustrates that the shell method handles scenarios where more intuitive methods like the disk method require complex transformations. The integration helps us evaluate which conditions will provide our solid a finite volume. Hence, the shell method becomes invaluable in varied problem scenarios of volumes of solids of revolution.
Improper Integrals
When dealing with calculations that extend to infinity, improper integrals become essential. These integrals help evaluate infinite behavior of functions, especially in calculus problems like volumes of solids of revolution.
An improper integral appears when limits of integration or the integrand itself are infinite.
For instance, the disk method resulted in \( V = \pi \int_1^{\infty} x^{-2p} \, dx \).
Here, the upper limit of the integral goes to infinity. To determine if this volume is finite, we check the convergence of the integral.The key to evaluating improper integrals is using the rule of power integrals: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \).
This rule guides us to examine if the resulting expression is finite as \( x \to \infty \). Convergence indicates a finite volume. For practical scenarios, this involves:
An improper integral appears when limits of integration or the integrand itself are infinite.
For instance, the disk method resulted in \( V = \pi \int_1^{\infty} x^{-2p} \, dx \).
Here, the upper limit of the integral goes to infinity. To determine if this volume is finite, we check the convergence of the integral.The key to evaluating improper integrals is using the rule of power integrals: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \).
This rule guides us to examine if the resulting expression is finite as \( x \to \infty \). Convergence indicates a finite volume. For practical scenarios, this involves:
- Determining convergence for the disk method, finite for \( p > 1 \)
- Determining convergence for the shell method, finite for \( p > 3 \)
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