Problem 14 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. \(y=\sqrt{4-2 x^{2}}, y=0,\) and \(x=0,\) in the first quadrant

Step-by-Step Solution

Answer

Question: Find the volume of the solid generated by revolving the region R bounded by the curves \(y = \sqrt{4 - 2x^2}, y=0,\) and \(x=0\) about the y-axis in the first quadrant using the shell method. Answer: The volume of the solid generated when the region R is revolved about the y-axis is \(\frac{16\pi}{3}\) cubic units.

Step 1: Solve for x as a function of y

We're given the function \(y = \sqrt{4 - 2x^2}\). We want to find x as a function of y. In order to do so, square both sides of the equation to eliminate the square root: \(y^2 = 4 - 2x^2.\) Now, rearrange the equation to solve for x: \(x^2 = \frac{4 - y^2}{2}.\) Our x-function is: \(x(y) = \sqrt{\frac{4 - y^2}{2}}.\)

Step 2: Determine limits of integration

To find our limits of integration, a and b, we need to find where the curve intersects the y-axis. We know the curve starts at y=0 and extends into the first quadrant, so a = 0. To find the end of the curve, we look for the value of y when x=0. When x = 0, our equation is \(y = \sqrt{4 - 2(0)^2}\), which simplifies to \(y = \sqrt{4}\). So, b = 2. We now have our limits of integration: 0 to 2.

Step 3: Write the shell method formula

Now that we have our x-function and our limits of integration, we can write the integral for the shell method formula: \(V = 2\pi\int_0^2 r(y)h(y)dy\), where r(y) is x(y) and h(y) is also x(y).

Step 4: Substitute x(y) into the formula

Substitute the x-function into the formula: \(V = 2\pi\int_0^2 (\sqrt{\frac{4 - y^2}{2}})(\sqrt{\frac{4 - y^2}{2}})dy.\) Simplify the integral: \(V = 2\pi\int_0^2 \frac{4 - y^2}{2} dy.\)

Step 5: Integrate and find the volume

Now, integrate the function with respect to y: \(V = 2\pi\left[\frac{1}{2}(4y - \frac{1}{3}y^3)\right]_0^2\). Evaluate the expression at the limits of integration: \(V = 2\pi\left(\frac{1}{2}(8 - \frac{8}{3}) - \frac{1}{2}(0 - 0)\right)\). Simplify and find the volume: \(V = 2\pi\left(\frac{8}{3}\right)\), \(V = \frac{16\pi}{3}\). The volume of the solid generated when the region R is revolved about the y-axis is \(\frac{16\pi}{3}\) cubic units.

Key Concepts

Shell Method

The shell method is a fantastic technique for finding volumes of solids of revolution. This method is particularly useful when revolving a region around an axis that doesn't touch the region, such as revolving around the y-axis when the function is in terms of y. To apply this method, imagine slicing the solid into thin cylindrical shells.

Here's the basic idea:

• Think of each slice as a hollow cylinder.
• The axis of revolution will be the center line of these cylinders.
• Calculate the volume of each cylindrical shell and add them up using integration.

For the region described in the exercise, we used the function in terms of y, with the shell formula for volume:\[ V = 2\pi \int_a^b r(y) h(y) \, dy \]where \( r(y) \) is the radius (distance from the axis of revolution to the shell), and \( h(y) \) is the height of the shell.

This method is very visual and tactile, making it easy to visualize how volume accumulates. Once you get a feel for the setup, integrating finds the sum of all volumes as the height of each shell varies.

Integration

Integration is the mathematical process of adding up infinitesimally small quantities to find an overall whole. In the context of volume calculation, integration helps us to sum up all the small slices or shells to form a complete solid. It's like summing up a series of thin layers to get the full volume.

In the given problem, after setting up the shell method integral, the process involved simplifying and solving:

• Total volume comes from integrating the thickness of each shell across the entire bound of integration.
• This involves simplifying the integral expression and evaluating it over specified limits.
• The integral in the solution: \[ \int_0^2 \frac{4 - y^2}{2} \, dy \], simplifies and solves to find the volume as \( V = \frac{16\pi}{3} \) cubic units.

Using integration, we translate a complicated geometric problem into an analytical calculation, leveraging calculus to yield the end result.

Mastering integration is essential for tackling numerous problems across math and science, especially when calculating areas, volumes, and many physical phenomena.

Revolution About the Axis

The idea of revolution about an axis is fundamental when considering problems of solids of revolution, a common scenario in calculus and engineering applications.

When a plane region rotates around an axis, it produces a three-dimensional shape. This process is what we explored when dealing with the region defined by the curves given in the exercise.

• Start with a defined region, which lies flat on a plane.
• Imagine this region "pivots" or "revolves" around the y-axis to generate a volume.
• The revolution moves every point in the region along a circular path, forming a solid.

With the given constraints of the functions and axes, we can decide the appropriate approach to quantify the new volume formed post-revolution.

The generated solid in this exercise is effectively symmetrical around the axis. We capture the resulting volume using integral calculus, which allows us to process this rotation systematically within our chosen coordinate system.