Problem 26 - Essential Calculus

Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(y\) -axis. $$ y=9-x^{2}, \quad y=0, \quad x=2, \quad x=3 $$

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region bounded by the given equations about the y-axis is \(21\pi\) units

Step 1: Identify the limits of integration

From the given bounds, observe that the region is from \(x = 2\) to \(x = 3\). Therefore, these are the limits of integration.

Step 2: Set up the integral

The volume \(V\) of the solid of revolution generated by rotating the region bounded by \(y=9-x^{2}\) and \(y=0\) about the \(y\)-axis can be found using the formula for the disk method: \(V = \pi \int_{a}^{b} r^{2} dx\), where \(r\) is the radius of the disc (the distance from the disc to the axis of rotation, in this case \(y\)), and \(a\) and \(b\) are the limits of integration. Here, \(r = \sqrt{9-x^{2}}\). Thus, \(V = \pi \int_{2}^{3} (\sqrt{9 - x^{2}})^{2} dx\). Simplify the integrand to obtain \(V = \pi \int_{2}^{3} (9 - x^{2}) dx\).

Step 3: Evaluate the integral

Perform the integral: \(V = \pi [9x - \frac{x^{3}}{3}]\_{2}^{3}\).

Step 4: Calculate the volume

Substitute the limits into the expression to find the volume: \(V = \pi (9*3 - \frac{3^{3}}{3}) - \pi (9*2 - \frac{2^{3}}{3})\). Simplify this to get \(V =\pi(21)\).

Key Concepts

Disk Method

The Disk Method is a powerful technique used in calculus to find the volume of a solid of revolution. This method is particularly useful when dealing with solids generated by rotating a region around an axis, such as the y-axis in our exercise.
Imagine slicing the solid into many thin disks perpendicular to the axis of rotation. Each disk's volume can be approximated, and adding up these volumes provides the total volume of the solid. The formula for the volume using the Disk Method is:

• \(V = \pi \int_{a}^{b} r^2 \, dx\)

In this formula, \(r\) is the function representing the distance from the axis of rotation to the outer edge of the disk (the radius).
In our example, the radius \(r\) is given by the square root of the function \(9 - x^2\). This comes from rewriting the equation \(y = 9 - x^2\), which describes the shape of the region. After setting up the integral, it simplifies to the expression using \(9 - x^2\) as the integrand.

Definite Integration

Definite Integration is key in calculating volumes through methods like the Disk Method. When you compute a definite integral, you're finding the total accumulation between the boundaries of your function. This is essential when determining the volume of a solid. Here, the process involves setting up an integral with specified boundaries (or limits), which are typically the values where the region of interest begins and ends. The integration itself results in a numerical answer that precisely represents the volume or area under the curve within those limits. In our problem, the integration is set up as:

• \(\int_{2}^{3} (9 - x^2) \, dx\)

This means we're interested in the accumulation of areas (which will sum up to a volume in this context) from \(x=2\) to \(x=3\). After integrating, any constant like \(\pi\) is applied to give the total volume.

Integration with Limits

Integration with Limits refers to the process of evaluating a definite integral across a bounded interval. In our example, the limits of integration are crucial because they determine the segment of the x-axis over which the function \(9 - x^2\) is integrated to find the volume.
The limits are directly sourced from the problem statement, telling us to focus between \(x=2\) and \(x=3\). These values define where our solid starts and ends along the x-axis. This bounded scope is important, as it ensures the calculated volume pertains only to the specified region, and not beyond.
Once you solve the integral:

• \(\left[9x - \frac{x^3}{3} \right]_{2}^{3}\)

You substitute the limits into the antiderivative, calculating two separate values for the upper limit (3) and the lower limit (2), subsequently finding the difference to get the total volume of the solid. This yields the final answer \(\pi(21)\) after all computations and simplifications are performed.