Problem 25 - 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. $$ y=x^{2}+3, x=1, x=2 $$

Step-by-Step Solution

Answer

The volume is \( \frac{33\pi}{2} \).

Step 1: Understand the Region Boundaries

The region we need to consider is bounded by the curve \( y = x^2 + 3 \) and the vertical lines \( x = 1 \) and \( x = 2 \). These boundaries define the area to be rotated around the \( y \)-axis.

Step 2: Setup for Volume by Cylindrical Shells

Since the rotation is around the \( y \)-axis, we can use the method of cylindrical shells. The volume \( V \) of the solid obtained by rotating a region around the \( y \)-axis is given by the integral:\[ V = 2\pi \int_{a}^{b} x \, f(x) \, dx \]where \( f(x) = x^2 + 3 \), and the limits of integration \( a = 1 \) and \( b = 2 \).

Step 3: Formulate the Integral Expression

Plug the expression for \( f(x) \) into the formula for volume:\[ V = 2\pi \int_{1}^{2} x(x^2 + 3) \, dx \]This simplifies to:\[ V = 2\pi \int_{1}^{2} (x^3 + 3x) \, dx \]

Step 4: Compute the Integral

Evaluate the integral term by term:\[ V = 2\pi \left[ \frac{x^4}{4} + \frac{3x^2}{2} \right]_{1}^{2} \]First, compute at \( x = 2 \):\[ \left( \frac{2^4}{4} + \frac{3(2^2)}{2} \right) = \left( 4 + 6 \right) = 10 \]Then compute at \( x = 1 \):\[ \left( \frac{1^4}{4} + \frac{3(1^2)}{2} \right) = \left( \frac{1}{4} + \frac{3}{2} \right) = \frac{7}{4} \]

Step 5: Calculate the Volume

Subtract the two integrals calculated:\[ V = 2\pi (10 - \frac{7}{4}) = 2\pi \left( \frac{40}{4} - \frac{7}{4} \right) = 2\pi \left( \frac{33}{4} \right) \]Finally, simplify the expression:\[ V = \frac{66\pi}{4} = \frac{33\pi}{2} \]

Step 6: Final Result

The volume of the solid formed by rotating the region bounded by \( y = x^2 + 3 \), \( x = 1 \), and \( x = 2 \) around the \( y \)-axis is \( \frac{33\pi}{2} \).

Key Concepts

Cylindrical Shells Method

The Cylindrical Shells Method is a powerful technique used to find the volume of a solid of revolution, particularly when the solid is generated by rotating an area around an axis that is not the axis of the function.
When dealing with a function of \( x \), such as \( y = x^2 + 3 \), and the rotation is around the \( y \)-axis, cylindrical shells provide an intuitive way to break down the problem.
The region is divided into thin vertical strips, which are considered as cylinders, or shells, when rotated. The volume of each shell is calculated, and then integrated across the defined range, from \( a \) to \( b \).

• The "height" of each cylindrical shell equals to the function value \( f(x) \).
• The "radius" of the shell is given by \( x \), the distance from the axis of rotation.
• The "thickness" of each shell is an infinitesimally small change in \( x \), denoted by \( dx \).

Using these definitions, we can express the volume \( V \), using the formula:\[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \]This integral sums up the volumes of all the individual shells from \( x = a \) to \( x = b \), giving us the total volume of the solid.

Definite Integral

A Definite Integral is a fundamental concept in calculus used for calculating the area under a curve or the total accumulation of quantities.
In the context of finding volumes through integration, it allows us to mathematically sum up infinitely small slices, or in the case of cylindrical shells, infinitely small cylindrical volumes.
The formula \( \int_{a}^{b} f(x) \, dx \) represents the process of taking the continuous sum of the function \( f(x) \) evaluated over the interval from \( a \) to \( b \). This gives us the area under the curve from \( a \) to \( b \).In problems involving volume, the definite integral is used to calculate the total volume of the rotational solid by integrating the expression derived for each cylindrical shell across the specified interval.
When you integrate an expression like \( x(x^2 + 3) \) over the interval from \( 1 \) to \( 2 \), as in the original exercise, you are accumulating the volume contributed by each shell over the entire range. This results in the total volume of the entire solid.

Rotation around the y-axis

When handling problems of Volume of Solids of Revolution where the rotation axis is the \( y \)-axis, it's key to adapt the approach based on this axis of rotation.
Typically we use the cylindrical shells method because it's more suitable than the disk method in these scenarios, where the region to be rotated is bounded vertically (i.e., by lines \( x = ext{constant} \)).
It involves seeing the solid as composed of multiple cylindrical sheets, or shells, that are centered on the \( y \)-axis and extend outward, parallel to the \( x \)-axis.

• The height of these shells is determined by the function (here, \( x^2 + 3 \)).
• The radius of each shell corresponds to \( x \), the variable of integration.

By "unrolling" each cylindrical shell, you can visualize it as a flat sheet of height equaling the function value and width determined by the circumference \( 2\pi x \), where \( x \) is the radius.
Rotating around the \( y \)-axis helps in easily integrating --- you simply take each infinitesimally thin shell, calculate its volume, and add up these volumes from the start of the interval to the end (from \( x=1 \) to \( x=2 \) in our exercise). Integrating this enables us to compute the full volume of the solid.