Problem 27 - 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. $$\text { The region enclosed by } x=\sqrt{5} y^{2}, \quad x=0, \quad y=-1, \quad y=1$$

Step-by-Step Solution

Answer

The volume is \(2\pi\).

Step 1: Identify the Region

The region is bounded by the curve \(x=\sqrt{5} y^{2}\), the vertical line \(x=0\), and the horizontal lines \(y=-1\) and \(y=1\). This is a parabolic region symmetric about the \(x\)-axis between \(y=-1\) and \(y=1\).

Step 2: Set Up the Integral Formula

For revolving around the \(y\)-axis, we use the disk method. The volume \(V\) is given by \[V = \, \pi \, \int_{-1}^{1} (f(y))^2 \, dy\]where \(f(y) = \sqrt{5} y^2\).

Step 3: Plug in the Function

Substitute \(f(y) = \sqrt{5} y^2\) into the integral. \[V = \pi \, \int_{-1}^{1} (\sqrt{5} y^2)^2 \, dy \] Simplifying the integrand gives: \[V = \pi \, \int_{-1}^{1} 5y^4 \, dy\]

Step 4: Integrate the Function

Calculate the integral: \[V = \pi [\frac{5}{5}y^5]_{-1}^{1} = \pi [y^5]_{-1}^{1} \] Evaluating the limits gives: \[V = \pi [(1)^5 - (-1)^5] = \pi [1 + 1] = 2\pi\]

Step 5: Conclude the Calculation

Thus, the volume of the solid generated by revolving the region around the \(y\)-axis is \(2\pi\).

Key Concepts

Volume of Solids

Calculating the volume of solids can be visualized like stacking thin slices, or disks, of an object. Imagine having a big block of cheese and slicing it into thin pieces. By stacking these on one another, we can find the total volume. In mathematics, we use similar principles to determine the volume of more complex shapes.

Solids of revolution are a common scenario where volumes are calculated. When a 2D shape is revolved about an axis, it creates a 3D solid. The interesting part is figuring out how to determine the volume of this solid thoroughly. Thankfully, calculus provides us tools to handle these problems efficiently.

When regions are rotated around an axis, the volume can be summed (or integrated) by adding up an infinite number of infinitely thin disks or washers. This technique is central to many volume calculation problems in calculus.

Key points to remember include:

• Revolution axis: The line about which the region is revolved to create the solid.
• Disk thickness: Tiny slices or disks, each with a small, consistent thickness.
• Integration bounds: The range over which the initial region stretches, often linked with limits in integration.

By understanding these key concepts, you can determine the volume of many interesting forms.

Disk Method

The disk method is a technique in calculus used to find the volume of a solid of revolution. When a region is revolved around the axis, each infinitesimal slice taken perpendicular to this axis forms a disk.

Consider a circular disk. It has a radius, and its volume can be thought of as the volume of each disk added together. Each small disk has a thickness and is similar to stacking up short cylinders to find a complete volume. The approach involves using an integral to sum up these small volumes into one total.

• Formula: For a solid revolved around the y-axis, the volume V is given by \[ V = \pi \int_{a}^{b} [f(y)]^2 \, dy \], where \(f(y)\) is the function describing the outer radius of the disks.
• Function substitution: Substituting the function into the formula to represent the radius in terms of the revolving axis.
• Integration: Evaluating this integral over the desired limits will yield the total volume.

This technique is particularly useful for shapes that are radially symmetrical or when slices are easily visualized as circles. By mastering the disk method, you can effectively determine the volume for a variety of revolved solids.

Definite Integrals

Definite integrals are a core concept in calculus used to calculate precise quantities, such as area, displacement, and volume. In the context of the disk method, definite integrals help determine the precise volume by adding up all of the infinitesimal contributions across an interval.

To compute a definite integral, one must combine the power of integration with specific bounds. These bounds are essential limits that define the start and endpoint of the area or volume under consideration.

Key characteristics of definite integrals include:

• Integration limits: The values at which the integral begins (lower limit) and ends (upper limit), which in the given problem are from -1 to 1.
• Evaluating antiderivatives: Finding the primitive function and then evaluating this at the boundaries to find the effect over the entire interval.
• Result interpretation: The result of a definite integral gives a finite value, here representing the total volume.