Problem 30 - Thomas Calculus in SI Units

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(y\) -axis. $$\begin{aligned}&\text { The region enclosed by } x=\sqrt{\cos (\pi y / 4)}, \quad-2 \leq y \leq 0\\\&x=0 \end{aligned}$$

Step-by-Step Solution

Answer

The volume of the solid is 4 cubic units.

Step 1: Understand the Region

The given curves are: \( x = \sqrt{\cos(\frac{\pi y}{4})} \) and \( x = 0 \), which defines a vertical boundary (the y-axis). We are asked to find the volume of the solid formed when this region is revolved around the y-axis. The region of interest is from \( y = -2 \) to \( y = 0 \).

Step 2: Set up the Integral

The volume of a solid of revolution around the y-axis can be found using the disk method. The formula is \( V = \pi \int_{c}^{d} [R(y)]^2 \, dy \), where \( R(y) \) is the radius function. Here, \( R(y) = \sqrt{\cos(\frac{\pi y}{4})} \). So, set up the integral: \[ V = \pi \int_{-2}^{0} \left(\sqrt{\cos(\frac{\pi y}{4})}\right)^2 \, dy \].

Step 3: Simplify the Integral

The integrand \( (\sqrt{\cos(\frac{\pi y}{4})})^2 \) simplifies to \( \cos(\frac{\pi y}{4}) \). This changes the integral to: \[ V = \pi \int_{-2}^{0} \cos(\frac{\pi y}{4}) \, dy \].

Step 4: Integrate the Function

Find the antiderivative of \( \cos(\frac{\pi y}{4}) \). Let \( u = \frac{\pi y}{4} \) and \( dy = \frac{4}{\pi} du \). The integral becomes: \[ V = \pi \times \frac{4}{\pi} \int \cos(u) \, du = 4 \int \cos(u) \, du = 4 \sin(u) + C. \].

Step 5: Evaluate the Definite Integral

Substitute back \( u = \frac{\pi y}{4} \) to get \( 4 \sin(\frac{\pi y}{4}) \). Evaluate the definite integral from \( y = -2 \) to \( y = 0 \):\[ V = 4[\sin(\frac{\pi \times 0}{4}) - \sin(\frac{\pi \times (-2)}{4})] = 4[0 - (-1)] = 4. \]

Step 6: State the Result

The volume of the solid generated by revolving the region bounded by the curves around the y-axis is 4 cubic units.

Key Concepts

Disk Method

The disk method is a technique used to find the volume of a solid of revolution. This solid is created when a region in the plane is revolved around a line, usually an axis like the x-axis or y-axis. This method involves integration and is especially handy when the region is bounded by a curve and a line, forming a shape that can be thought of as many tiny disks stacked along the axis of rotation.

When using the disk method, the integral is set up using the formula:

• \[ V = \pi \int_{a}^{b} [R(x)]^2 \, dx \]

where \( R(x) \) refers to the radius of each disk, and \([a, b]\) denotes the interval of integration, along the axis of rotation. For revolutions around the y-axis, you would use \( R(y) \) and integrate with respect to \( y \).

This method transforms the volume problem into an area under the curve problem, allowing us to compute the volume by essentially summing the areas of concentric circles scaled across the region.

Antiderivative of Trigonometric Functions

Antiderivatives play a crucial role in solving integrals, particularly when handling functions like trigonometric functions. In our example, we dealt with the function \( \cos\left(\frac{\pi y}{4}\right) \). To find its antiderivative, a substitution method can often be employed to simplify the expression.

For \( \cos\left(\frac{\pi y}{4}\right) \), we set \( u = \frac{\pi y}{4} \), which simplifies the differential to \( dy = \frac{4}{\pi} du \). This makes the integration straightforward.

• The antiderivative of \( \cos(u) \) is \( \sin(u) \) plus a constant.
• Thus, the integral \( \int \cos(u) \, du = \sin(u) + C \).
• Don’t forget to substitute back \( u \) to the original variable \( y \) to finalize the solution.

Knowing how to find antiderivatives of trigonometric functions and simplify them through substitution is key in tackling integrals involving these functions.

Definite Integrals

A definite integral is used to calculate the total area underneath a curve within a specific range. It's a powerful tool for determining quantities like area, distance, and, in this context, volume of solids.

In our problem, we use definite integrals to find the volume by revolving a region around an axis. The process involves:

• Setting up an integral with a specific lower and upper bound, defined by the particular function and range.
• Integrating over this range to find the exact value, in this case, using \( y = -2 \) to \( y = 0 \).
• Once the antiderivative is computed, substitute the limits back into the antiderivative to evaluate the integral.

The definite integral enables us to compute the exact volume of the solid created by revolving the region. It's like summing infinitely many, infinitely small quantities to reach the total volume.

Bounded Regions

Bounded regions play a central role in determining the limits of integration when calculating areas or volumes. These regions are defined by specific curves or lines serving as boundaries which encapsulate the shape.

Understanding how to identify and interpret these boundaries is crucial. In the exercise provided:

• The region is bounded by the curve \( x = \sqrt{\cos(\frac{\pi y}{4})} \) and the line \( x = 0 \).
• The bounds for the integral are determined by the intersection points or specified limits, here \( y = -2 \) and \( y = 0 \).
• Such regions can lie above, below, or between other curves or lines, necessitating careful analysis of where they intersect or enclose.

Knowing exactly where these boundaries lie helps in setting up the right integral and accurately calculating the desired volume.