Problem 32 - Thomas Calculus

Question

The region bounded by \(y=2 x-x^{2}\) and \(y=x\) about a. the \(y\) -axis b. the line \(x=1\)

Step-by-Step Solution

Answer

Both volumes are \(\frac{\pi}{6}\).

Step 1: Find Points of Intersection

To determine the region bounded by the curves, we need to find their points of intersection by equating the equations of the curves: \(2x - x^2 = x\). Simplifying gives us \(-x^2 + x = 0\), which factors to \(x(x - 1) = 0\). Thus, the points of intersection are \(x = 0\) and \(x = 1\).

Step 2: Set up the Integral For Rotation About the y-axis

For part (a), we will use the method of cylindrical shells to rotate the region around the \(y\)-axis. The formula for the volume of a solid of revolution using cylindrical shells is \(V = \int_{a}^{b} 2\pi x (f(x) - g(x)) \, dx\). Here, \(f(x) = 2x - x^2\) and \(g(x) = x\).

Step 3: Calculate the Volume For y-axis

Apply the shell method from \(x = 0\) to \(x = 1\). The volume \(V\) is given by \[V = \int_{0}^{1} 2\pi x ((2x - x^2) - x) \, dx = \int_{0}^{1} 2\pi x (x - x^2) \, dx.\] Simplifying inside the integral gives \[V = \int_{0}^{1} 2\pi x^2 - 2\pi x^3 \, dx.\]Integrating, we have \[V = 2\pi \left[\frac{x^3}{3} - \frac{x^4}{4} \right]_0^1.\]Evaluating, \[V = 2\pi \left(\frac{1}{3} - \frac{1}{4} \right) = 2\pi \left(\frac{4}{12} - \frac{3}{12}\right) = \frac{2\pi}{12} = \frac{\pi}{6}.\]

Step 4: Set up the Integral For Rotation About x=1

For part (b), we will use the method of cylindrical shells to rotate about the line \(x=1\). The formula now becomes \(V = \int_{a}^{b} 2\pi (1-x) (f(x) - g(x)) \, dx\), where \(f(x) = 2x - x^2\) and \(g(x) = x\).

Step 5: Calculate the Volume For x=1

Set up the integral from \(x = 0\) to \(x = 1\): \[V = \int_{0}^{1} 2\pi (1-x) ((2x - x^2) - x) \, dx = \int_{0}^{1} 2\pi (1-x) (x - x^2) \, dx.\]Simplifying the expression gives \[V = \int_{0}^{1} 2\pi ((1-x)x - (1-x)x^2) \, dx = \int_{0}^{1} 2\pi (x - x^2 - x^2 + x^3) \, dx = \int_{0}^{1} 2\pi (x - 2x^2 + x^3) \, dx.\]Integrating term by term,\[V = 2\pi \left[\frac{x^2}{2} - \frac{2x^3}{3} + \frac{x^4}{4} \right]_0^1.\]Evaluating gives \[V = 2\pi \left(\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\right).\]Finding a common denominator, \[V = 2\pi \left(\frac{6}{12} - \frac{8}{12} + \frac{3}{12}\right) = 2\pi \left(\frac{1}{12}\right) = \frac{\pi}{6}.\]

Step 6: Summary Conclusion

The volumes of the solids of revolution about both the \(y\)-axis and the line \(x=1\) are both \(\frac{\pi}{6}\).

Key Concepts

Cylindrical Shells

The method of cylindrical shells is a powerful technique in calculus that simplifies the process of finding the volume of a solid of revolution. When a region is rotated about a vertical axis, like the y-axis, we form cylindrical layers as opposing to a pile of disks. Each cylindrical shell's volume is structured by "wrapping" a rectangle around an axis, essentially resembling a label around a can.

The formula for the volume of a solid using this method is:
\[ V = \int_{a}^{b} 2\pi x (f(x) - g(x)) \, dx \] where:

• \(x\) is the distance from the axis of rotation
• \(f(x)\) and \(g(x)\) are the functions that create the top and bottom of the shell

As we see, it spaces integrating along x, making it excel when rotating around vertical lines, like the y-axis in our problem.

Solid of Revolution

A solid of revolution arises when a plane figure is revolved around a line—known as the axis of revolution. This process creates a three-dimensional object. When we define regions using functions and then revolve around a line, these regions generate patterns similar to cups, vases, or other round shapes.

By revolving the region bounded by \(y=2x-x^2\) and \(y=x\), about different axes like \(y\)-axis or a line \(x=1\), we produce unique three-dimensional shapes. Using techniques like cylindrical shells, we calculate the volumes within these fascinating solids by considering the sum of these tiny volumes.

Integration

Integration plays a crucial role in finding the volume of solids of revolution. It allows us to determine the sum of infinitely many small quantities, namely the individual cylindrical shells or disks. In the context of using cylindrical shells, integration sums up the areas of all shells along the x-axis or other directed limits.

For integrating in this exercise, the steps included:

• Setting up the integral to calculate the volume about an axis
• Simplifying the expressions inside the integral
• Executing the definite integral from the lower to upper limits

Integrals give us an effective way of encapsulating the effects of the infinitesimally small elements needed to find areas and volumes, a vital component in calculus.

Points of Intersection

Before we even begin rotating or finding volumes, we find it crucial to determine where two curves meet—called their points of intersection. These points help us understand the bounds for our calculations, effectively marking out the region to be rotated.

In this example, we equate the equations \(y = 2x - x^2\) and \(y = x\), resulting in a simple zero equation \(-x^2 + x = 0\). Solving this gives us solutions \(x = 0\) and \(x = 1\).

With these intersection points, we gather a clear picture of the region to be manipulated, defining the boundary limits for our integrals and ensuring accurate calculations.