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 \(x\) -axis. \(y=\sqrt{50-2 x^{2}},\) in the first quadrant

Short answer

Based on the given information and using the disk method, the volume of the solid generated when the region R is revolved about the x-axis is calculated to be π.

Step-by-step solution

1. Identify the given region, axis of rotation, and boundaries

The problem has already provided the functions that form the given region \(R\). We know that it is bounded by the curves \(y=\frac{1}{\sqrt[4]{1-x^{2}}}\), \(y=0\), \(x=-\frac{1}{2}\), and \(x=\frac{1}{2}\). The axis of rotation is the \(x\)-axis.

2. Configure the function in terms of the distance to the axis of rotation

We will revolve our region around the x-axis, so we need a function to represent the distance from the curve to the x-axis (at any given x-value). Since \(y=\frac{1}{\sqrt[4]{1-x^{2}}}\), we have \(r(y) = \frac{1}{\sqrt[4]{1-x^{2}}}\).

3. Determine the limits of integration

The limits of integration are given as \(x=-\frac{1}{2}\) and \(x=\frac{1}{2}\), which represent the left and right boundaries of our region, respectively.

4. Apply the disk method formula

Using the disk method formula, we have to compute the integral of the area of each infinitesimal disk from the left boundary (\(x = -\frac{1}{2}\)) to the right boundary (\(x = \frac{1}{2}\)). The volume \(V\) of the solid is given by: $$V = \int \pi r^2 \, dx = \int_{-\frac{1}{2}}^{\frac{1}{2}} \pi\large(\frac{1}{\sqrt[4]{1-x^{2}}}\large)^2\, dx$$

5. Compute the integral

Now we have to compute the integral: \begin{align*} V &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \pi\large(\frac{1}{\sqrt[4]{1-x^{2}}}\large)^2\, dx\\ &=\pi\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{1}{\sqrt{1-x^{2}}}\, dx \end{align*} The integral can be evaluated using the substitution \(x=\sin u\), $$ dx=\cos u \, du $$ So the integral becomes \begin{align*} V &= \pi\int_{-\arcsin(\frac{1}{2})}^{\arcsin(\frac{1}{2})} \frac{1}{\sqrt{1-\sin^2 u}}\cos{u} \, du \\ &=\pi\int_{-\arcsin(\frac{1}{2})}^{\arcsin(\frac{1}{2})} \cos u \, du \end{align*} Now, we can integrate \(\cos u\) with respect to \(u\): $$V = \pi\left[\sin u\right]_{-\arcsin(\frac{1}{2})}^{\arcsin(\frac{1}{2})}$$ Finally, plug in the limits of integration: $$V=\pi\left(\sin(\arcsin\frac{1}{2}) - \sin(-\arcsin\frac{1}{2})\right) = \pi\left(\frac{1}{2}+\frac{1}{2}\right) = \pi$$ So, the volume of the solid generated when \(R\) is revolved about the x-axis is \(\pi\).

Key concepts

Volume of Solids of Revolution

When you revolve a region around an axis, a three-dimensional solid is formed. The disk method is one way to find the volume of such a solid. Imagine slicing the solid into thin disks or circles perpendicular to the axis of rotation. Each disk is essentially a very thin cylinder, and by adding up the volume of every tiny disk from one end to the other, you get the total volume of the solid. Here, the function that defines the region is revolved around the x-axis. The formula for the volume, using the disk method, is:

• Volume, \( V = \int \pi r^2 \, dx \)

where \(r\) is the distance from the curve to the axis of rotation (i.e., the radius of the disk).
The limits of integration are defined by the boundaries of the region being revolved.

The curve \(y = \frac{1}{\sqrt[4]{1-x^2}}\) is rotated, so the radius function (\(r\)) becomes this same function in terms of \(x\).

Definite Integration

Definite integration helps find the accumulated quantity, like area or volume, between a set range. In this problem, it is the process of adding up the infinitesimally small disks along the x-axis to find the total volume.
When you set up a definite integral:

• You use limits to show where to start and stop the calculation. These limits are \(x = -\frac{1}{2}\) and \(x = \frac{1}{2}\) for this exercise.
• You integrate the function \(\pi \left(\frac{1}{\sqrt[4]{1-x^2}}\right)^2\). This involves evaluating how the function accumulates values over the interval.

The result of a definite integral, along these bounds, provides us with the volume of the solid.

Substitution Method

Sometimes, direct integration of a function can be complex. The substitution method is a technique that simplifies the integral by introducing a new variable. In this case, we use substitution to integrate the function more easily.
To solve this integral, we let:

• \( x = \sin u \), so \( dx = \cos u \, du \)
• This simplifies the integral to \( \int \pi \cos u \, du \)
This change of variables transforms the function, making it easier to perform the integration. After the substitution, you integrate with respect to \(u\), and substitute back to get back to \(x\), ensuring it fits within the original limits by converting them into \(u\) space:
• The limits change from \(-\arcsin(\frac{1}{2})\) to \(\arcsin(\frac{1}{2})\).
Finally, the calculated definite integral gives us the volume of \( \pi \).