Problem 2
Question
In Problems 1-12, find the volume of the solid generated when the region \(R\) bounded by the given curves is revolved about the indicated axis. Do this by performing the following steps. (a) Sketch the region \(R\). (b) Show a typical rectangular slice properly labeled. (c) Write a formula for the approximate volume of the shell generated by this slice. (d) Set up the corresponding integral. (e) Evaluate this integral. \(y=x^{2}, x=1, y=0 ;\) about the \(y\)-axis
Step-by-Step Solution
Verified Answer
The volume of the solid is \( \frac{\pi}{2} \).
1Step 1: Sketch the Region R
The region \( R \) is bounded by the curves \( y = x^2 \), \( x = 1 \), and \( y = 0 \). To sketch this region, draw the parabolic curve \( y = x^2 \), the vertical line \( x = 1 \), and the horizontal line \( y = 0 \). The region \( R \) is the area under the parabola from \( x = 0 \) to \( x = 1 \), above the \( y = 0 \) line.
2Step 2: Label a Typical Rectangular Slice
Choose a thin vertical slice at position \( x \) with a small width \( \Delta x \). This slice is a rectangle bounded by the curve \( y = x^2 \) above, \( y = 0 \) below, \( x \) on the left, and \( x + \Delta x \) on the right.
3Step 3: Formula for the Volume of the Shell
When revolved around the \( y \)-axis, each rectangular slice generates a cylindrical shell. The height of the shell is \( y = x^2 \), the radius is \( x \), and the thickness is \( \Delta x \). Thus, the volume of the shell is approximately \( 2\pi x (x^2) \Delta x = 2\pi x^3 \Delta x \).
4Step 4: Set Up the Integral for Total Volume
Integrate the approximate volume of the cylindrical shell over the interval from \( x = 0 \) to \( x = 1 \). The integral for the total volume \( V \) is: \[V = \int_{0}^{1} 2\pi x^3 \, dx\]
5Step 5: Evaluate the Integral
Calculate the integral: \[V = 2\pi \int_{0}^{1} x^3 \, dx\]Find the antiderivative of \( x^3 \), which is \( \frac{x^4}{4} \), and evaluate from 0 to 1: \[V = 2\pi \left[ \frac{x^4}{4} \right]_{0}^{1} = 2\pi \left( \frac{1^4}{4} - \frac{0^4}{4} \right) = 2\pi \left( \frac{1}{4} \right) = \frac{\pi}{2}\]
6Step 6: Conclude the Problem
Thus, the volume of the solid generated when the region \( R \) is revolved about the \( y \)-axis is \( \frac{\pi}{2} \).
Key Concepts
Cylindrical Shells MethodIntegral CalculusAntiderivativesDefinite Integrals
Cylindrical Shells Method
The cylindrical shells method is a very handy tool in integral calculus for finding the volume of solids of revolution. Imagine taking a thin vertical strip from the region that you want to revolve. When you revolve this strip around an axis, it forms a cylindrical shell. That's where this method gets its name.
Here are the key parts involved in this method: - **Radius**: The radius of the cylindrical shell is the distance from the axis of rotation. In the exercise above, the axis is the y-axis, so the radius for a slice at position \( x \) is simply \( x \). - **Height**: The height of the shell is given by the function that bounds the region. In our case, this is \( y = x^2 \). - **Thickness**: The thickness of the shell is infinitesimally small and is denoted by \( \Delta x \), representing an infinitely small change in \( x \).
The volume of a thin shell is approximated as \( 2\pi \times \text{radius} \times \text{height} \times \text{thickness} \). So, for our exercise, each shell has a volume of \( 2\pi x^3 \Delta x \). By summing these shells from \( x = 0 \) to \( x = 1 \) using an integral, we can calculate the total volume.
Here are the key parts involved in this method: - **Radius**: The radius of the cylindrical shell is the distance from the axis of rotation. In the exercise above, the axis is the y-axis, so the radius for a slice at position \( x \) is simply \( x \). - **Height**: The height of the shell is given by the function that bounds the region. In our case, this is \( y = x^2 \). - **Thickness**: The thickness of the shell is infinitesimally small and is denoted by \( \Delta x \), representing an infinitely small change in \( x \).
The volume of a thin shell is approximated as \( 2\pi \times \text{radius} \times \text{height} \times \text{thickness} \). So, for our exercise, each shell has a volume of \( 2\pi x^3 \Delta x \). By summing these shells from \( x = 0 \) to \( x = 1 \) using an integral, we can calculate the total volume.
Integral Calculus
Integral calculus is fundamentally about accumulation, which in this context, refers to summing infinite small pieces to find a whole, like an area or volume. It's like adding up the piled-up layers to see the big picture.
When dealing with physical problems, you can think of the integral as a super clever sum. You are summing up infinite little bits (like the cylindrical shells) to build up to a total quantity.
In the problem we're examining, we use integral calculus to accumulate the volumes of infinitesimally thin shells from \( x = 0 \) to \( x = 1 \). The basic calculus tool here is the integral:\[V = \int_{0}^{1} 2\pi x^3 \, dx\]This integral tells us the whole volume by adding up all those tiny shell volumes, effectively gluing them together to form the solid.
When dealing with physical problems, you can think of the integral as a super clever sum. You are summing up infinite little bits (like the cylindrical shells) to build up to a total quantity.
In the problem we're examining, we use integral calculus to accumulate the volumes of infinitesimally thin shells from \( x = 0 \) to \( x = 1 \). The basic calculus tool here is the integral:\[V = \int_{0}^{1} 2\pi x^3 \, dx\]This integral tells us the whole volume by adding up all those tiny shell volumes, effectively gluing them together to form the solid.
Antiderivatives
An antiderivative is about reversing the process of differentiation. It's finding a function whose derivative gives us back the original function. In simple terms, if you know how fast something is changing, an antiderivative helps you figure out where it is (or was).
In this problem, to find the volume, we need the antiderivative of \( x^3 \). Let's see why that’s key: - The original function \( x^3 \) represents the changes happening across the range for the shells.- Its antiderivative is \( \frac{x^4}{4} \). Achieving this involves using one of the fundamental rules of antiderivatives: for \( x^n \), add 1 to the power and divide by the new power, which gives us the height of each accumulated slice when \( x \) varies over \([0, 1]\).
This step is crucial because it anchors the remainder of our steps in the integral process. Without it, we'd be unable to accurately compute the total volume.
In this problem, to find the volume, we need the antiderivative of \( x^3 \). Let's see why that’s key: - The original function \( x^3 \) represents the changes happening across the range for the shells.- Its antiderivative is \( \frac{x^4}{4} \). Achieving this involves using one of the fundamental rules of antiderivatives: for \( x^n \), add 1 to the power and divide by the new power, which gives us the height of each accumulated slice when \( x \) varies over \([0, 1]\).
This step is crucial because it anchors the remainder of our steps in the integral process. Without it, we'd be unable to accurately compute the total volume.
Definite Integrals
Definite integrals are like the icing on the cake in integral calculus. They allow you to evaluate the accumulation of quantities over a specific range. For example, how much total volume is there from where \( x = 0 \) to where \( x = 1 \)?
In our problem, once we have the antiderivative, we move towards calculating the definite integral. This involves evaluating the antiderivative at the upper limit and subtracting the value at the lower limit: - To finish up the integral \( \int_{0}^{1} 2\pi x^3 \, dx \), we evaluated \( 2\pi \left[ \frac{x^4}{4} \right]_{0}^{1} \).- Plug in the upper limit first (1) into \( \frac{x^4}{4} \) and then the lower limit (0).
This gives us \( 2\pi \left( \frac{1^4}{4} \right) - 2\pi \left( \frac{0^4}{4} \right) \), which simplifies nicely to \( \frac{\pi}{2} \). Thus, with the definite integral, we not only understand how things accumulate but also get precise values over a specified interval.
In our problem, once we have the antiderivative, we move towards calculating the definite integral. This involves evaluating the antiderivative at the upper limit and subtracting the value at the lower limit: - To finish up the integral \( \int_{0}^{1} 2\pi x^3 \, dx \), we evaluated \( 2\pi \left[ \frac{x^4}{4} \right]_{0}^{1} \).- Plug in the upper limit first (1) into \( \frac{x^4}{4} \) and then the lower limit (0).
This gives us \( 2\pi \left( \frac{1^4}{4} \right) - 2\pi \left( \frac{0^4}{4} \right) \), which simplifies nicely to \( \frac{\pi}{2} \). Thus, with the definite integral, we not only understand how things accumulate but also get precise values over a specified interval.
Other exercises in this chapter
Problem 1
A discrete probability distribution for a random variable \(X\) is given. Use the given distribution to find (a) \(P(X \geq 2)\) and (b) \(E(X)\). \begin{array}
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A discrete probability distribution for a random variable \(X\) is given. Use the given distribution to find (a) \(P(X \geq 2)\) and (b) \(E(X)\). $$ \begin{arr
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