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Chapter 6: Problem 49

Find the volume of the solid generated by revolving each region about the \(y\) -axis. The region in the first quadrant bounded above by the parabola \(y=x^{2},\) below by the \(x\) -axis, and on the right by the line \(x=2\)

Short Answer

Expert verified
The volume is \(\frac{16\pi}{3}\).

Step by step solution

01

Identify the Problem

We are tasked with finding the volume of a solid formed by revolving a region around the y-axis. The region is in the first quadrant, bounded by the parabola \(y = x^2\), the x-axis, and the line \(x = 2\).
02

Express x as a function of y

Since we are revolving around the y-axis and the original function is \(y = x^2\), we need to express \(x\) as a function of \(y\). From \(y = x^2\), solving for \(x\) gives us \(x = \sqrt{y}\).
03

Define the Region of Integration

The region of interest is between the x-axis (\(y = 0\)) and where the parabola meets the line \(x = 2\). At \(x = 2\), we substitute into the parabola equation: \(y = 2^2 = 4\). Thus, our integration bounds are from \(y = 0\) to \(y = 4\).
04

Set Up the Integral for Volume

We use the method of cylindrical shells to find the volume. The formula for the volume of a solid of revolution using shells is: \[ V = 2\pi \int_{a}^{b} x f(x) \, dy \]. In terms of y, the expression is \(x = \sqrt{y}\), and radius \(x\) as \(x = x\). The correct integral is thus: \[ V = 2\pi \int_{0}^{4} \sqrt{y} \, (2 - \sqrt{y}) \, dy \].
05

Solve the Integral

Expand the integrand: \(\sqrt{y} (2 - \sqrt{y}) = 2\sqrt{y} - y\). This gives the integral \[ V = 2\pi \int_{0}^{4} (2\sqrt{y} - y) \, dy \]. Now, compute: - The integral of \(2\sqrt{y}\) is \(\frac{4}{3} y^{3/2}\) - The integral of \(y\) is \(\frac{1}{2} y^2\) Thus, the full integral to solve is: \[ V = 2\pi \left[ \frac{4}{3} y^{3/2} - \frac{1}{2} y^2 \right]_0^4 \].
06

Evaluate the Integral

Compute the definite integral: 1. At \(y = 4\): \[ \frac{4}{3} (4)^{3/2} - \frac{1}{2} (4)^2 = \frac{4}{3} (8) - \frac{1}{2} (16) = \frac{32}{3} - 8 \] 2. At \(y = 0\): The expression evaluates to 0. The result is: \[ \frac{32}{3} - 8 = \frac{32}{3} - \frac{24}{3} = \frac{8}{3} \].
07

Compute the Final Volume

Multiply by the factor outside the integral: \[ V = 2\pi \left( \frac{8}{3} \right) = \frac{16\pi}{3} \].
08

Conclusion

The volume of the solid generated by revolving the given region around the y-axis is \(\frac{16\pi}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Shells Method
The Cylindrical Shells Method is a powerful technique used to find the volume of solids of revolution. When we revolve a 2D region around an axis, this method involves envisioning the solid as made up of many thin, hollow cylinders (or shells). Each shell contributes to the total volume of the solid.
Here's how it works:
  • Imagine slicing the solid vertically into thin slices parallel to the axis of rotation.
  • Each slice, or "shell," resembles a cylinder with a small thickness.
  • The volume of each shell is calculated by multiplying its circumference, height, and thickness.
  • For a shell revolving around the y-axis, the thickness is representative of an infinitesimally small change in the y-values (hence, dy).
In our example, the curve from which we derive the cylindrical shells is defined by the bounds of the region that needs to be rotated around the y-axis. This ensures the shells correctly form around the axis and the enclosed volume reflects the original shape.
Volume Calculation
The essence of finding volume using the Cylindrical Shells Method lies in setting up a correct integral. In our example, we revolve a region bounded by the parabola \(y = x^2\), the x-axis, and the line \(x = 2\) around the y-axis.
The formula used to compute the volume is:\[ V = 2\pi \, \int_{a}^{b} x f(x) \, dy \]where:
  • \(2\pi\) accounts for the rotation around the axis, generating the shell's circumference,
  • \(x\) represents the radius from the axis of revolution, and
  • \(f(x)\) is the height of the shell parallel to the axis of revolution.
In the context of our problem, we express \(x\) in terms of \(y\), yielding \(x = \sqrt{y} \). This helps calculate the height and boundaries of the cylindrical shells for integration.
Integration Bounds
Determining the right integration bounds is crucial. These bounds define the interval over which the integral is computed. They encompass the region that is being revolved around the axis.
For our exercise:
  • The bounds come from the intersection of the region's defining limits with the axis of revolution.
  • The bottom of the region is defined by the x-axis or \(y = 0\).
  • The top limit is where the parabola \(y = x^2\) intersects the vertical boundary line \(x = 2\), which is calculated by substituting \(x = 2\) into the parabola equation, giving \(y = 4\).
Thus, the integral is evaluated from \(y = 0\) to \(y = 4\), encompassing the full height of the region that is being revolved.
Revolving Around y-axis
When a region is revolved around the y-axis, it forms a three-dimensional solid. This revolution turns vertical lines in the region into circular shells, effectively sweeping through space to build a volume.
Steps to visualize this:
  • Choose the region on a graph bounded by given functions or lines.
  • Revolve each point in the vertical slice (line) around the y-axis.
  • Every revolution traces out a circle, creating the shell鈥檚 thickness in the y-direction.
This method provides an intuitive approach to understanding how flat areas can be spun to form recognizable 3D shapes. For the given problem, revolving the region between the parabola, x-axis, and line \(x=2\) creates our desired solid, where each "slice" of the parabola gives rise to a unique shell contributing to the overall volume.

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Most popular questions from this chapter

Find the areas of the surfaces generated by revolving the curves about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. $$x=2 \sqrt{4}-y, \quad 0 \leq y \leq 15 / 4 ; \quad y-\text { axis }$$

Use a CAS to perform the following steps for the given graph of the function over the closed interval. a. Plot the curve together with the polygonal path approximations for \(n=2,4,8\) partition points over the interval. (See Figure \(6.22 .)\) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for \(n=2.4 .8\) with the actual length given by the integral. How does the actual length compare with the approximations as \(n\) increases? Explain your answer. $$f(x)=x^{2} \cos x, \quad 0 \leq x \leq \pi$$

Find the center of mass of a thin plate of constant density \(\delta\) covering the given region. The region bounded by the parabola \(y=25-x^{2}\) and the \(x\) -axis

a. Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis. b. Graph the curve to see what it looks like. If you can, graph the surface too. c. Use your utility's integral evaluator to find the surface's area numerically. \(x=\sin y, \quad 0 \leq y \leq \pi ; \quad y\) -axis

The graph of \(y=x^{2}\) on \(0 \leq x \leq 2\) is revolved about the \(y\) -axis to form a tank that is then filled with salt water from the Dead Sea (weighing approximately 73 lb/ft \(^{3}\) ). How much work does it take to pump all of the water to the top of the tank?
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