Problem 10 - Thomas Calculus

Question

Use the shell method to find the volumes of the solids generated by revolving the regions bounded by the curves and lines in Exercises \(7-12\) about the \(y\) -axis. $$ y=2-x^{2}, \quad y=x^{2}, \quad x=0 $$

Step-by-Step Solution

Answer

The volume is \( \pi \).

Step 1: Identify the Region of Interest

First, identify the area bounded by the given curves and lines. The curves are \( y = 2-x^2 \) and \( y = x^2 \), and the line is \( x = 0 \). The region is where the curves intersect and are bounded by \( x = 0 \).

Step 2: Find Points of Intersection

To find the points of intersection of \( y = 2-x^2 \) and \( y = x^2 \), set them equal to each other: \( 2-x^2 = x^2 \). Solving, we get \( 2 = 2x^2 \) leading to \( x^2 = 1 \), so \( x = \pm 1 \). However, since we are bounded by \( x = 0 \), we only consider \( x = 0 \) and \( x = 1 \).

Step 3: Set up the Shell Method Integral

Using the shell method, the volume \( V \) of the solid is given by \[ V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dx \] where the height is the vertical distance between the curves, \( 2-x^2 - x^2 = 2-2x^2 \), and the radius is the distance from the \( y\)-axis, which is \( x \). Thus, the integral becomes \[ V = 2\pi \int_{0}^{1} x(2-2x^2) \, dx \].

Step 4: Evaluate the Integral

Calculate the integral: first distribute to get \[ \int_{0}^{1} x(2-2x^2) \, dx = \int_{0}^{1} (2x - 2x^3) \, dx \]. Separate the integrals: \[ \int_{0}^{1} 2x \, dx - \int_{0}^{1} 2x^3 \, dx \]. Integrate to get \[ 2\left[\frac{x^2}{2}\right]_{0}^{1} - 2\left[\frac{x^4}{4}\right]_{0}^{1} \]. This simplifies to \[ [1 - 0] - [0.5 - 0] = 1 - 0.5 = 0.5 \].

Step 5: Calculate Final Volume

Multiply the result by \( 2\pi \): \[ V = 2\pi \times 0.5 = \pi \]. Thus, the volume of the solid is \( \pi \).

Key Concepts

Volumes of Solids of Revolution

The shell method is a technique used to find the volume of solids formed by rotating a region around a specific axis. When we revolve a bounded region around an axis, we create a 3D solid called a "solid of revolution." This is particularly important in calculus for calculating volumes that are not easily measurable through traditional geometric formulas. The shell method specifically involves using cylindrical shells (hence its name) to approximate and then sum up the volume of the solid.
The formula for the shell method is \[ V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dx \]This formula calculates the volume by integrating the volume of the cylindrical shells over the interval \[a, b\]. Here:

• "Radius" refers to the perpendicular distance from the axis of rotation to the shell.
• "Height" stands for the height of the shell at a given point.

The result of this integration gives us the total volume of the solid. The shell method is particularly beneficial when dealing with solids rotated around the y-axis, as it simplifies the process significantly.

Curves Intersection

Finding where curves intersect is crucial when dealing with bounded regions. These intersections give us the limits of integration, as they define the limits above or below which the area exists. Often, this involves solving equations to find the x-values where two curves meet, which will serve as the boundaries for your calculus operations. In this case, the curves \( y = 2-x^2 \) and \( y = x^2 \) intersect when they equal each other.
We solve for the intersection by setting the two equations equal: \[ 2-x^2 = x^2 \] This simplifies to \[ 2 = 2x^2 \]which further reduces to \[ x^2 = 1 \].

• Thus, the x-values at which the curves intersect are \( x = \pm 1 \). However, we limit our attention to \( x = 0 \) and \( x = 1 \) since the region is also bounded by the line \( x = 0 \).

Overall, intersection points play a significant role in setting up the limits for our integrals and defining the bounded region needed for volume calculations.

Integral Evaluation

Integral evaluation is the core part of finding volumes using the shell method. After setting up the shell method integral formula, the next step is to carry out the integral evaluation to compute the exact volume.
Here, our integral is \[ \int_{0}^{1} x(2-2x^2) \, dx \]To solve this, we start by distributing \( x \) into \( 2-2x^2 \) to simplify it to \[ \int_{0}^{1} (2x - 2x^3) \, dx \].

• This results in two separate integrals:
• \( \int_{0}^{1} 2x \, dx \) which evaluates to \( [x^2] \) from 0 to 1 and results in 1.
• \( -\int_{0}^{1} 2x^3 \, dx \) which evaluates to \( [0.5x^4] \) and results in 0.5.

The final calculation of assembling these yields 1 - 0.5 = 0.5. This value is then used in a final multiplication with \( 2\pi \), resulting in the final volume \( V = \pi \). Breaking down integrals into manageable parts is vital in ensuring that each component is accurately computed.

Bounded Regions

A bounded region in calculus refers to an area enclosed by curves and lines, within which we wish to compute a property like area or volume. When dealing with rotating regions to find solids of revolution, understanding the bounded region is particularly critical as it defines where the area of interest is located.
For our exercise, the bounded region is enclosed by the curves \( y = 2-x^2 \) and \( y = x^2 \) and the line \( x = 0 \):

• The intersection of the curves gives x-boundaries that are essential for setting up limits in the integral.
• The line \( x = 0 \) further defines the region, ensuring we are only considering the volume generated by rotating the specific area between \( x = 0 \) and \( x = 1 \).

Understanding bounded regions enables us to correctly set up the integral limits and ensure the accuracy of the resulting volume measurement. It helps to visualize or graph the region to make these boundaries clearer when solving.