Problem 27 - Calculus and its applications

Question

Find the volume generated by rotating the area bounded by the graphs of each set of equations around the \(y\) -axis. $$ y=\sqrt{x}, x=4, x=9 $$

Step-by-Step Solution

Answer

The volume is \( 125.19\pi \) cubic units.

Step 1: Set Up the Problem

We begin by identifying the area to be rotated. The area is bounded by the curve \( y = \sqrt{x} \), the vertical lines \( x = 4 \) and \( x = 9 \), and the x-axis. We will rotate this region about the y-axis.

Step 2: Express \(x\) in Terms of \(y\)

Since we are rotating around the \(y\)-axis, we solve for \(x\) in terms of \(y\). From \(y = \sqrt{x}\), we have \(x = y^2\). The bounds for \(y\) are from \(\sqrt{4} = 2\) to \(\sqrt{9} = 3\).

Step 3: Setup the Integral using the Washer Method

The formula for the volume of a solid of revolution using the washer method around the \(y\)-axis is \( V = \pi \int_{a}^{b} [R(y)^2 - r(y)^2] \, dy\). For this problem, \( R(y) = 9 \) and \( r(y) = y^2 \). The limits of integration are from \( 2 \) to \( 3 \).

Step 4: Compute the Volume Integral

Compute the integral\[V = \pi \int_{2}^{3} [(9)^2 - (y^2)^2] \, dy = \pi \int_{2}^{3} [81 - y^4] \, dy\]. Calculate it as follows:\[= \pi \left[ 81y - \frac{y^5}{5} \right]_2^3\].

Step 5: Evaluate the Integral Result

Substitute the limits into the integrated function:\[V = \pi \left( (81 \times 3 - \frac{3^5}{5}) - (81 \times 2 - \frac{2^5}{5}) \right)\].Perform the calculations:\[= \pi \left( 243 - \frac{243}{5} - 162 + \frac{32}{5} \right)\]\[= \pi \left( 243 - 48.6 - 162 + 6.4 \right)\]\[= \pi (39.8)\]\[\approx 125.19\].

Step 6: Conclusion

Thus, the volume of the solid obtained by rotating the region bounded by \(y = \sqrt{x}, x = 4,\) and \( x = 9 \) about the \(y\)-axis is \( 125.19\pi \) cubic units.

Key Concepts

Washer Method

The **Washer Method** is a technique used in calculus to find the volume of a solid of revolution. This method applies when there is a hollow region or hole in the solid; essentially, it calculates the volume of a "washed out" disk or annulus. It's particularly useful when the solid is generated by rotating an area around an axis that creates a shape with an outer and inner radius.When using the Washer Method, we generally follow these steps:

• Identify the region of interest and the axis of rotation.
• Determine the expressions for the outer radius, \(R(y)\), and the inner radius, \(r(y)\) in terms of the variable of integration, which often depends on the axis of rotation.
• Set up and compute the definite integral using the formula: \[V = \pi \int_{a}^{b} [R(y)^2 - r(y)^2] \, dy\]where \(a\) and \(b\) are the integration limits for \(y\).
• Evaluate the integral to find the volume.

In our specific exercise, we rotated the region bounded by the graphs around the \(y\)-axis, which led us to the expression of the outer radius being constant at "9" and the inner radius being \(y^2\). The computation of the integral then gives the total volume, illustrating the washer approach efficiently.

Definite Integrals

**Definite Integrals** are a fundamental part of calculus, especially in applications involving areas and volumes. They provide a way to calculate the total accumulation of quantities, like area or volume under certain conditions, over an interval.In the context of solids of revolution and volume calculations:

• Definite integrals are used to sum up small slices or disks of area, which together form the volume of the solid.
• They involve integrating a function over a specific, finite interval identified by the bounds \(a\) and \(b\).
• The result of evaluating a definite integral provides the exact measurement of the desired quantity, such as volume.

In our example, the definite integral was set up with limits from \(y = 2\) to \(y = 3\) and incorporated expressions for the squared radii to calculate the solid's volume. The result was a precise measurement in cubic units, specifically \(125.19\pi\), showcasing the power of definite integrals in practical applications.

Solid of Revolution

A **Solid of Revolution** is a three-dimensional object created by rotating a two-dimensional region around an axis. This technique is common in calculus to generate and analyze real-world objects that have rotational symmetry.To visualize how a solid of revolution forms:

• Imagine the given area, defined by curves or lines, as a sheet in the plane.
• When this sheet rotates about an axis, every point in the plane revolves, tracing out a complete, symmetrical shape in 3D space.
• The resulting object has cross-sections that are circular disks or washers, depending on whether the "sheet" starts with gaps or holes relative to the axis.

In essence, understanding solids of revolution opens the door to a deeper grasp of geometric concepts and their applications in physical science and engineering. For the exercise at hand, the region bounded by \(y = \sqrt{x}\), \(x = 4\), and \(x = 9\) was rotated around the \(y\)-axis to form the solid. This showcases a classic example of a solid of revolution derived from a relatively simple bounded area.