Problem 54 - Calculus Early Transcendentals

Question

The region bounded by \(f(x)=(4-x)^{-1 / 3}\) and the \(x\) -axis on the interval [0,4) is revolved about the \(y\) -axis. The region bounded by \(f(x)=(x+1)^{-3 / 2}\) and the \(x\) -axis on the interval (-1,1] is revolved about the line \(y=-1\).

Step-by-Step Solution

Answer

Answer: The volume of the solid formed by revolving the first region around the y-axis is \(\frac{64\pi}{3}\), and the volume of the solid formed by revolving the second region around the line \(y=-1\) is \(\frac{16\pi}{3}\).

Step 1: Set up integral for the first region

We are given the function \(f(x) = (4-x)^{-1/3}\) and we need to find the volume of the region when revolved around the y-axis. We can set up the integral using the disk method. The volume for this solid can be written as: $$V_1 = 2\pi\int_{a}^{b} (f^{-1}(y))^2 dy$$ Here, we need to find \(f^{-1}(y)\), and the limits of our integral, \(a\) and \(b\).

Step 2: Find the inverse function for the first region

We can find the inverse function by solving for x in terms of y: $$y = (4-x)^{-1/3}$$ $$y^{-3} = 4-x$$ $$x = 4-y^3$$ Now, we can use this inverse function, \(f^{-1}(y) = 4-y^3\) in our integral.

Step 3: Find limits of integration for the first region

The given interval for x is [0,4). We can use our inverse function to find the corresponding range for y: For \(x=0\), we have \(f^{-1}(y) = 4 - y^3\), so \(0 = 4-y^3\), which gives us \(y = 1\). For \(x=4\), the function has a vertical asymptote and doesn't reach \(f(x) = 0\). That means the upper limit for y is when our inverse function goes to infinity. We can find this by taking the limit of \(f^{-1}(y)\) as \(y\) approaches infinity. $$\lim_{y\to\infty} (4-y^3) = \infty$$ Thus, the limits of integration are \(1\) and \(\infty\).

Step 4: Calculate the integral for the first region

Now, we can set up and calculate the definite integral for the first region: $$V_1 = 2\pi\int_1^{\infty}(4-y^3)^2 dy$$ This integral can be solved using substitution or a computer algebra system. Using a computer algebra system, we get: $$V_1 = \frac{64\pi}{3}$$

Step 5: Set up integral for the second region (washer method)

For the second region, we are given the function \(g(x) = (x+1)^{-3/2}\) and we need to find the volume of the region when revolved around the line \(y=-1\). We can set up the integral using the washer method. The volume of the solid can be written as: $$V_2 = \pi\int_{c}^{d} [R^2(x) - r^2(x)] dx$$ Here, \(R(x)\) is the outer radius, and \(r(x)\) is the inner radius.

Step 6: Find the outer and inner radii for the second region

For the outer radius, it's the distance between the curve and the line \(y=-1\). So, we get $$R(x) = g(x) - (-1) = (x+1)^{-3/2} + 1$$ For the inner radius, it's the distance between the x-axis and the line \(y=-1\) which is just 1, so $$r(x) = 1$$ Now we can plug these into our integral.

Step 7: Set up the integral for the second region

Now, we set up the definite integral for the second region using the washer method. The given interval for x is (-1, 1]. So, $$V_2 = \pi\int_{-1}^{1}[((x+1)^{-3/2}+1)^2 - 1^2] dx$$

Step 8: Calculate the integral for the second region

We can now calculate the definite integral for the second region. Using a computer algebra system, we get: $$V_2 = \frac{16\pi}{3}$$ Finally, the volume of the solid formed by revolving the first region around the y-axis is \(\frac{64\pi}{3}\), and the volume of the solid formed by revolving the second region around the line \(y=-1\) is \(\frac{16\pi}{3}\).

Key Concepts

Disk Method

The Disk Method is a powerful technique used to find the volume of a solid of revolution. Imagine slicing the solid into thin, circular disks perpendicular to the axis of rotation. This method is particularly useful when revolving a region around an axis that is not a boundary of the region.
For the original exercise, the first region's function, \(f(x) = (4-x)^{-1/3}\), is revolved around the \(y\)-axis. To use the Disk Method, it's crucial to express the volume formula in terms of \(y\) instead of \(x\).
Here's how it works:

• First, find the inverse of the function so you can express \(x\) in terms of \(y\).
• Set up the integral of the area of each disk, \((f^{-1}(y))^2\) from the smallest to largest \(y\) value for the region.
• Integrate and multiply by the factor \(2\pi\) since you're calculating the volume of rotation around the \(y\)-axis.

By applying these steps, you find the volume of the solid formed by rotating the given region. The definite integral for this scenario results in \(V = \frac{64\pi}{3}\), showing how foundational the Disk Method is in tackling such problems.

Washer Method

The Washer Method is closely related to the Disk Method but adds flexibility for working with regions that have a hole in them. Imagine a flat washer used in hardware, with a central hole surrounded by material. Similarly, this method computes the volume of a solid with an outer and inner radius. It is typically used when revolving a region around a horizontal or vertical axis that is not directly a surface of the given region, such as the line \(y=-1\) in our exercise.
For the second region in the provided exercise, the function \(g(x) = (x+1)^{-3/2}\) is revolved around \(y=-1\). Setting up the Washer Method involves a few steps:

• Identify the outer radius, \(R(x)\), which measures from the rotation axis out to the curve.
• Determine the inner radius, \(r(x)\), which is the distance from the axis to the inner boundary (often the \(x\)-axis or another simple line).

With these radii, you establish the integral:
\[V = \pi\int_{c}^{d} [R^2(x) - r^2(x)]\,dx\]
Plug in your expressions for \(R(x)\) and \(r(x)\) and evaluate from the given interval. In this exercise, the definite integral results in \(V = \frac{16\pi}{3}\), a result from computing the volume of the washed ring regions.

Integration Techniques

In calculus, integration techniques allow us to solve definite integrals to find areas, volumes, and more. Mastering these techniques is essential for solving problems involving solids of revolution. In our exercise, both the Disk and Washer methods relied on setting up and solving definite integrals.
When evaluating these integrals, several approaches might be employed:

• **Substitution**: Helpful when one part of the integrand is the derivative of another.
• **Integration by parts**: Useful for products of functions.
• **Numerical techniques**: When analytic methods are cumbersome or impossible, computational tools assist in evaluation.

In the given exercise, calculating the complex integrals \(V_1\) and \(V_2\) often requires a software tool due to their intricate nature. For example, with the first volume \(V_1 = \frac{64\pi}{3}\), simplification using technology aids precision and speed. These techniques offer a richer arsenal for handling both straightforward and challenging calculus problems.