Problem 31 - Thomas Calculus

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(y\)-axis. 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

The volume of the solid is \( 2\pi \).

Step 1: Set Up the Integral for the Volume

We're using the disk method for revolving around the y-axis. The formula for the volume of a solid of revolution around the y-axis is \[ V = \pi \int_{a}^{b} [R(y)]^2 \, dy \]where \( R(y) \) is the distance from the y-axis to the curve. Our curve is described by \( x = \sqrt{2\sin(2y)} \).

Step 2: Identify the Limits of Integration and the Outer Radius

The region is bounded between \( y = 0 \) and \( y = \pi/2 \). The outer radius \( R(y) \) is the function \( R(y) = \sqrt{2\sin(2y)} \) based on the problem specification of \( x = \sqrt{2\sin(2y)} \).

Step 3: Write the Integral with the Given Boundaries

Substituting \( R(y) = \sqrt{2\sin(2y)} \) into the volume formula, we get:\[ V = \pi \int_{0}^{\pi/2} (\sqrt{2\sin(2y)})^2 \, dy \]. We simplify \((\sqrt{2\sin(2y)})^2\) to \(2\sin(2y)\).

Step 4: Evaluate the Integral

Solving the integral, we have:\[ V = \pi \int_{0}^{\pi/2} 2\sin(2y) \, dy \].First, find the antiderivative:\[ \int 2\sin(2y) \, dy = -\cos(2y) \].Evaluating from \( 0 \) to \( \pi/2 \), we get:\[ V = -\left[\cos(2y)\right]_{0}^{\pi/2} \].

Step 5: Calculate the Definite Integral Value

Substitute the limits into the antiderivative:\(-[\cos(\pi) - \cos(0)] \).We know \( \cos(\pi) = -1 \) and \( \cos(0) = 1 \), so:\[ -(-1 - 1) = 2 \].

Step 6: Compute the Final Volume

Multiply by \( \pi \):\[ V = \pi \times 2 = 2\pi \].Thus, the volume of the solid is \( 2\pi \).

Key Concepts

Disk Method

The Disk Method is a technique in Calculus used to find the volume of a solid of revolution. When you revolve a region around an axis, you form a three-dimensional shape. The Disk Method involves slicing this solid shape into thin, circular disks, perpendicular to the axis of rotation. Each of these disks has a small thickness and a certain radius.

• The formula to find the volume of a solid of revolution using the Disk Method is

\[ V = \pi \int_{a}^{b} [R(y)]^2 \, dy\]Here,

• \( V \) is the total volume of the solid,
• \( R(y) \) is the radius of each disk,
• and \( [a, b] \) are the limits of integration that define the region.

This method is particularly useful for revolving regions around the y-axis, as we just need to identify the radius as a function of \( y \). The disks are parallel to the axis of rotation, making calculations simpler. This concept is vital in finding volumes of complex solids where using basic geometric formulas is not possible.

Integral Calculus

Integral Calculus is a branch of Calculus focused on accumulation and area under curves, often used to compute areas, volumes, and central points. When used in the context of the Disk Method, it helps us calculate the sum of infinite small disks that create the whole solid. The process involves:

• Finding the integral of the function that describes the solid's cross-sectional area,
• Considering the limits of integration based on the boundaries of the region,
• And, evaluating this integral to get the volume.

The integration process essentially adds up the infinite disks, acknowledging how each tiny change contributes to the total volume. Integral calculus provides a framework to precisely and systematically handle these accumulations, which is beyond basic algebraic calculations. Understanding this helps in figuring out how parts come together to form the whole, with accuracy and mathematical rigor.

Definite Integrals

Definite Integrals are a specific type of integral used frequently in Calculus. Unlike indefinite integrals, they calculate the accumulation of quantities over a specified interval. For the Disk Method, definite integrals help determine the total volume of the solid by evaluating the integral from a starting point \( a \) to an endpoint \( b \).Key features of definite integrals include:

• Evaluation within specific bounds, making them applicable to real-world and geometric problems,
• Providing exact numerical results,
• And, being essential in computations involving rates of change and total accumulations.

In our exercise, after calculating the integral, we substituted the limits of \( y = 0 \) and \( y = \pi/2 \) to find the final volume. This process showcased how definite integrals help reduce a complex shape into a manageable calculation. By applying these principles, we obtained the exact volume, demonstrating the power of calculus in solving practical problems.