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Magazine Chapter 6: Problem 24
Find the volume of the solid that results when the region enclosed by the given curves is revolved about the \(y\) -axis. $$x=1-y^{2}, x=2+y^{2}, y=-1, y=1$$
Short Answer
Expert verified
The volume is \( \frac{32\pi}{15} \).
Step by step solution
01
Understand the Problem
We need to find the volume of the solid formed by revolving the region enclosed by the curves around the y-axis. The curves are given as \(x = 1-y^2\) and \(x = 2+y^2\), bounded between \(y = -1\) and \(y = 1\).
02
Sketch the Region
First, sketch the curves to visualize the enclosed region. The curve \(x = 1-y^2\) is a downward-opening parabola, while \(x = 2+y^2\) is an upward-opening parabola. These curves intersect at \(y = \pm 1\), within which the region is enclosed by these curves.
03
Set Up the Integral for Volume
When the region between these curves is revolved around the y-axis, we use the method of cylindrical shells. The formula for the volume using cylindrical shells is: \[ V = 2\pi \int_{a}^{b} (radius)(height) \, dy \] where the radius is \(|x|\) from the y-axis, and the height is the difference between the curves.
04
Identify Radius and Height
For a small horizontal strip at \(y\), the radius is the distance from the y-axis to the strip, given by \(x\). The height is the difference between the outer and inner curve, so height = \((2+y^2) - (1-y^2)\).
05
Simplify Height Expression
Calculate the height by simplifying: \[ height = (2+y^2) - (1-y^2) = 2y^2 + 1 \]
06
Determine Radius Range
We need the radius values for the cylindrical shells. Using the given curves, for \(y\) from -1 to +1, the outer curve \(x = 2+y^2\) gives the outermost radius and \(x = 1-y^2\) is the innermost radius.
07
Set up the Integral Limits and Solve
Set the bounds of integration from \(y = -1\) to \(y = 1\). The integral for the volume becomes:\[ V = 2\pi \int_{-1}^{1} (2+y^2) (2y^2+1) \, dy \] Perform the integration and calculate:1. Expand the integrand: \[ (2+y^2)(2y^2+1) = 4y^2 + 2 + 2y^4 + y^2 \]2. Simplify: \[ 2y^4 + 5y^2 + 2 \]3. Integrate term by term and compute.
08
Integrate and Compute
Perform the integration term by term:\[ \int_{-1}^{1} (2y^4 + 5y^2 + 2) \ dy = \left[ \frac{2}{5}y^5 + \frac{5}{3}y^3 + 2y \right]_{-1}^{1} \]Evaluate the definite integral by substituting limits:\[ \left( \frac{2}{5}(1)^5 + \frac{5}{3}(1)^3 + 2(1) \right) - \left( \frac{2}{5}(-1)^5 + \frac{5}{3}(-1)^3 + 2(-1) \right) \]Simplify the result to find the total volume.
09
Calculate Final Result
After evaluating the terms, subtract the results from \[ y = 1 \] and \[ y = -1 \]. Compute the arithmetic to obtain the final volume: \[ V = 2\pi \left( \frac{16}{15} \right) = \frac{32\pi}{15} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cylindrical Shells
When dealing with the volume of solids of revolution, the method of cylindrical shells is a powerful tool. It is particularly useful when the region is rotated around an axis parallel to the variable being integrated, like the y-axis in our exercise. The idea here is to consider the solid as composed of infinitely many thin cylindrical shells.
- Radius: The distance from the axis of rotation (y-axis) to a shell is represented as the radius of that shell. For a horizontal strip at any point y, this will be the x-coordinate value, here being each function value of x.
- Height: This is determined by the points of intersection of the curves. It is the difference between the values of the two curves at any y. In this problem, it is given by the expression \((2+y^2) - (1-y^2)\).
- A cylindrical shell is a thin slice of the solid made by rotating a vertical line segment. Its volume is given by the surface area (circumference times height) times thickness (dy).
Definite Integral
A definite integral allows us to calculate the exact value for a continuous quantity over a specified interval. In the context of calculating volumes, the definite integral helps to add up an infinite number of infinitesimal elements, such as disks or shells, to find a total volume.
Here鈥檚 the process you follow:
Here鈥檚 the process you follow:
- Determine the upper and lower bounds of integration, which correspond to the interval where the curves interact or enclose a region.
- The integral is formulated as \[ \int_{a}^{b} f(y)\,dy \], where \( f(y) \) is your function describing the part of the solid (like the height times radius for shells).
- The evaluation of the integral over the bounds gives the accumulated "sum" of volumes of these shells from the start to the end of the interval.
- Each individual contribution to the area under the integral adds up to form the total volume, leading to the profound concept that small parts can build up a complete whole.
Intersection of Curves
Finding where curves intersect is crucial for understanding which regions are enclosed and how they interact. This is essential for setting the correct bounds and accurately defining the formulas for heights or other elements in integral calculus.
In our given problem:
In our given problem:
- Setting Equations: First, we find where the curves intersect by solving \( 1-y^2 = 2+y^2 \). This equation simplifies to find practical points of intersection.
- Simplifying: Solving this directly gives intersection points at \( y = \pm 1 \). These points are where the curves cross each other vertically, indicating that between these points, the functions define an enclosed region.
- Resulting Region: Knowing these points allows us to properly set the limits for our integration and to see which curve is outer or inner relative to the axis of rotation.
