Problem 42 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Let \(S\) be the solid generated when \(R\) is revolved about the given axis. If possible, find the volume of \(S\) by both the disk/washer and shell methods. Check that your results agree and state which method is easiest to apply. \(y=x^{2} / 8, y=2-x,\) and \(x=0 ;\) revolved about the \(y\) -axis

Step-by-Step Solution

Answer

Question: Find the volume of the solid when the region R bounded by the curves \(y=\frac{x^2}{8}\), \(y=2-x\), and \(x=0\) is revolved around the y-axis. Confirm your answer using both the disc/washer and shell methods. Answer: The volume of the solid when the region R is revolved around the y-axis is \(8\pi\) cubic units. This result is confirmed using both the disc/washer and shell methods.

Step 1: Plot the curves and determine intersection points

Draw the curves on the same graph to visualize the region R enclosed by those curves. Find the points of intersection between the curves by solving the system of equations: $$ y = \frac{x^2}{8},\quad y = 2-x $$ Set the two equations equal to each other and solve for x: $$ \frac{x^2}{8} = 2-x $$ After solving the quadratic equation, we obtain x values of the intersection points: \(x = 2\) and \(x = 4\). Now, substitute these x values back into any of the equations to find the corresponding y values. The intersection points are \((2,0)\) and \((4, -2)\).

Step 2: Set up the integrals for the disc/washer method

To set up the integral for the disc/washer method, we will use the volume formula with cross-sectional areas. $$ V = \pi \int_{a}^{b} (R^2 - r^2) dy $$ In our case, the outer radius R is given by the line \(x=2\), and the inner radius r is given by \(x^2/8\). We can rewrite these as functions of y since we revolve the region around the y-axis. Outer radius function: $$ R(y) = 2 $$ Inner radius function: $$ x = (8y)^{1/2}\:\text{or}\: r(y) = \sqrt{8y} $$ The range of integration will be from \(y=0\) to \(y=2\) (the y values of the intersection points). $$ V = \pi \int_{0}^{2} \left(2^2 - (\sqrt{8y})^2 \right) dy $$

Step 3: Evaluate the disc/washer integral

Now that we have the integral set up, we can evaluate it. $$ V = \pi \int_{0}^{2} (4 - 8y) dy $$ This results in the following calculation after integrating and evaluating the limits: $$ V = \pi \left[4y - 4y^2\right]_{0}^{2} = \pi[8 - 16] = 8\pi $$ The volume of the solid S using the disc/washer method is \(8\pi\) cubic units.

Step 4: Set up the integral for the shell method

Next, we will use the shell method to find the volume of the solid. The shell method formula is as follows: $$ V = 2\pi \int_{a}^{b} (x * h(x)) dx $$ In our case, the height function \(h(x)\) will be the difference between the two functions of x, \(2-x\) and \(x^2/8\). The range of integration will be from \(x=0\) to \(x=4\). $$ V = 2\pi \int_{0}^{4} x[(2-x) - \frac{x^2}{8}] dx $$

Step 5: Evaluate the shell integral

Now we can evaluate the integral for the shell method. $$ V = 2\pi \int_{0}^{4} x(2-x-\frac{x^2}{8}) dx $$ The integration and evaluation of the limits result in the same volume value as before: $$ V = 8\pi $$ The volume of the solid S using the shell method is also \(8\pi\) cubic units. #Conclusion#: Both methods produce the same result of \(8\pi\). In this case, neither method is significantly more straightforward to apply than the other, as both methods involve simple integrals with clear geometric interpretations.

Key Concepts

Disk Method

The Disk Method is used to find the volume of a solid of revolution, where the solid is formed by rotating a region around an axis. This method involves slicing the solid perpendicular to the axis of rotation, forming disk-shaped cross-sections.
To calculate the volume using this method, we consider the circular disks. The formula to find the volume, depending on whether there is a hole or not, is:

• If there is no hole (pure disk): \[ V = \pi \int_{a}^{b} [R(y)]^2 \, dy \]Here, \( R(y) \) is the radius of the disk.
• If there is a central hole (washer): \[ V = \pi \int_{a}^{b} (R(y)^2 - r(y)^2) \, dy \]Here, \( R(y) \) is the outer radius, and \( r(y) \) is the inner radius.

For the given problem, the outer radius comes from the line \( x=2 \), while the inner radius is the curve \( y = \frac{x^2}{8} \), rewritten here as functions of \( y \) for integration around the \( y \)-axis. The calculation provides the volume after applying limits of integration from \( y=0 \) to \( y=2 \). This results in the volume \( 8\pi \) cubic units.

Shell Method

Unlike the Disk Method, the Shell Method involves revolving around the axis, where you consider cylindrical shells. It is often more straightforward for solids revolving around the vertical axis, typically the \( y \)-axis, when the region is described with functions of \( x \).
In this technique, the volume is calculated by summing up the volumes of thin cylindrical shells:

• Volume of shells: \[ V = 2\pi \int_{a}^{b} (x \cdot h(x)) \, dx \]
• Here, \( x \) is the radius of the shell, and \( h(x) \) is the height of the shell.

For our problem, \( h(x) \) is the difference between the functions \( 2-x \) and \( \frac{x^2}{8} \), representing the height of the region.
The integration occurs over the interval \( x=0 \) to \( x=4 \), which again results in the same volume \( 8\pi \) cubic units. Using this method can sometimes simplify the setup when dealing with the \( y \)-axis.

Intersection Points

Finding intersection points of curves is crucial when dealing with solids of revolution. It helps define boundaries or limits of integration.
In the given exercise, the curves are \( y = \frac{x^2}{8} \) and \( y = 2-x \). By setting these two equations equal, you can solve for \( x \) to find where they intersect:

• Solve: \[ \frac{x^2}{8} = 2 - x \].
• The solution to this gives the x-values of intersection: \( x=2 \) and \( x=4 \).

These points are then plugged back into the original equations to find corresponding y-values, resulting in points \((2,0)\) and \((4,-2)\). These intersection points are not only critical in setting up the integrals but also ensure that the region \( R \) is appropriately defined.

Definite Integrals

Definite Integrals play an essential role in calculating the volumes of solids of revolution. They provide a way to sum infinitesimally small quantities over a specific interval.
For a solid of revolution, these integrals are used to add up the tiny volumes of disk sections or cylindrical shells. Here’s why they’re important:

• They allow calculation over a given interval, ensuring the entire region is considered.
• They take into account both the radius and height in forming volumes, adapting to the shape of curves.

In the exercise, for both the Disk and Shell Methods, definite integrals help to find exact volumes by analyzing how functions intersect and form boundaries. By evaluating these integrals, either through manual calculations or numerical methods, the solid’s volume is accurately determined as \( 8\pi \). This shows how mathematical theory translates into practical volume calculations.