Problem 29 - 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. $$\text { The region enclosed by } x=\sqrt{2 \sin 2 y}, \quad 0 \leq y \leq \pi / 2, \quad x=0$$

Step-by-Step Solution

Answer

Volume is \(2\pi\).

Step 1: Understand the Problem

We are asked to find the volume of the solid obtained by revolving the given region around the y-axis. The region is bounded by the curve \(x = \sqrt{2 \sin 2y}\), the line \(x = 0\), the line \(y = 0\), and the line \(y = \frac{\pi}{2}\). We will use the method of disks (or washers) to calculate this volume.

Step 2: Set Up the Integral

The volume \(V\) of the solid formed by the revolution of the region about the \(y\)-axis is given by the integral formula:\[V = \pi \int_{a}^{b} [R(y)]^2 \, dy\]Here, \(R(y)\) is the radius of the disk at height \(y\), which is \(x = \sqrt{2 \sin 2y}\). Thus, the integral becomes:\[V = \pi \int_{0}^{\frac{\pi}{2}} (\sqrt{2 \sin 2y})^2 \, dy\]

Step 3: Simplify the Integral Expression

After substituting \(R(y)\) into the integral, we simplify:\[V = \pi \int_{0}^{\frac{\pi}{2}} 2 \sin 2y \, dy\]Thus, the integral we need to evaluate is:\[V = 2\pi \int_{0}^{\frac{\pi}{2}} \sin 2y \, dy\]

Step 4: Compute the Integral

To evaluate the integral, we use the substitution \(u = 2y\), \(du = 2 \, dy\), so \(dy = \frac{1}{2} \, du\) and change the limits of integration. When \(y=0\), \(u=0\); when \(y=\frac{\pi}{2}\), \(u=\pi\). Thus, the integral becomes:\[V = 2\pi \int_{0}^{\pi} \sin u \cdot \frac{1}{2} \, du = \pi \int_{0}^{\pi} \sin u \, du\]Integrating, we obtain:\[V = \pi [-\cos u]_{0}^{\pi} = \pi [-\cos(\pi) + \cos(0)] = \pi [1 + 1] = 2\pi\]

Step 5: Conclude the Solution

The integral has been computed, and we find that the volume of the solid generated by rotating the region about the \(y\)-axis is \(2\pi\).

Key Concepts

Disk Method

The Disk Method is a powerful technique in calculus used to find the volume of a solid of revolution. Imagine you take a flat region on the coordinate plane and spin it around an axis. The resulting shape is called a solid of revolution. This method slices the solid into a series of disk-like shapes perpendicular to the axis of rotation. It's like stacking pancakes, but with each pancake being a thin slice of the solid. To determine the volume of one of these disks, we first need to know the radius, which corresponds to the distance from the axis of rotation to the curve of interest. In the context of the problem, the distance or radius, denoted as \(R(y)\), is \(x = \sqrt{2 \sin 2y}\).

• The formula for the volume of a solid using the Disk Method is \(V = \pi \int_{a}^{b} [R(y)]^2 \, dy\), where \([a, b]\) are the bounds along the rotation axis.
• This integral sums up the volumes of all those disk slices to give the total volume of the solid.

By following the Disk Method, we transform a complex 3D shape into a manageable mathematical problem.

Definite Integral

The Definite Integral is a fundamental concept in calculus that calculates the exact area under a curve or the total accumulation of quantities. When you apply it to the Disk Method, it allows you to sum up all the infinitesimally small disks to find the total volume of a solid of revolution.In this exercise, we set up the integral with limits from \(0\) to \(\frac{\pi}{2}\), which correspond to the region's bounds on the \(y\)-axis. The integral expression \(\int_{0}^{\frac{\pi}{2}} 2 \sin 2y \, dy\) represents the sum of all the areas of the disk faces multiplied by a small thickness, effectively giving the volume when rotated around the axis.

• The definite integral allows for precise calculation of volumes, giving exact values instead of estimates.
• When evaluating a definite integral, we calculate the difference between the antiderivative values at the upper and lower bounds.

Using definite integrals in problems like these simplifies the process and ensures an accurate result.

Trigonometric Substitution

Trigonometric substitution is a technique utilized to evaluate integrals, especially when they include trigonometric functions. This substitution simplifies the integral by transforming complex trigonometric expressions into simpler ones.The key to using trigonometric substitution effectively is to identify the trigonometric identity that will simplify your work. For this exercise, we use the substitution \(u = 2y\), which simplifies the integration process. With this substitution:

• The differential \(du = 2 \, dy\) implies that \(dy = \frac{1}{2} \, du\).
• The limits of integration change as well: when \(y=0\), \(u=0\); when \(y=\frac{\pi}{2}\), \(u=\pi\).

Transforming the integral gives us: \(V = 2\pi \int_{0}^{\pi} \sin u \cdot \frac{1}{2} \, du = \pi \int_{0}^{\pi} \sin u \, du\).Once the integral is transformed, it becomes simpler to solve. The result, after evaluating the integral, is \(V = 2\pi\), which represents the volume of the revolved solid.