Problem 140 - Calculus Volume 1

Question

For the following exercises, find the volume generated when the region between the curves is rotated around the given axis. y = 3 ? x, y = 0, x = 0, and x = 2 rotated around the y-axis.

Step-by-Step Solution

Answer

The volume of the solid is \( \frac{16\pi\sqrt{6}}{5} \).

Step 1: Understanding the Problem

The task involves finding the volume of a solid of revolution formed by the region bounded by the curves and lines: \( y = \sqrt{3x} \), \( y = 0 \), \( x = 0 \), and \( x = 2 \) when it is rotated around the \( y \)-axis.

Step 2: Setting Up the Integral

To find the volume, we'll use the method of cylindrical shells. The formula for the volume of a solid of revolution using cylindrical shells is \[V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx,\] where \( f(x) \) is the function in terms of \( x \). Here, \( f(x) = \sqrt{3x} \), \( a = 0 \), and \( b = 2 \).

Step 3: Computing the Integral

Substitute the function and limits into the integral:\[V = 2\pi \int_{0}^{2} x \sqrt{3x} \, dx.\]First, rewrite \( \sqrt{3x} \) as \( (3x)^{1/2} \), and simplify:\[V = 2\pi \int_{0}^{2} x \cdot (3x)^{1/2} \, dx = 2\pi \int_{0}^{2} \sqrt{3} x^{3/2} \, dx.\]

Step 4: Integration Process

Perform the integration: Start by integrating \( \sqrt{3} x^{3/2} \):\[\int x^{3/2} \, dx = \frac{2}{5}x^{5/2} + C.\]So,\[V = 2\pi \left[ \sqrt{3} \cdot \frac{2}{5} x^{5/2} \right]_{0}^{2} = 2\pi \cdot \frac{2\sqrt{3}}{5} \left( x^{5/2} \right)_{0}^{2}.\]

Step 5: Evaluating the Definite Integral

Evaluate \( \frac{2\sqrt{3}}{5} x^{5/2} \) from 0 to 2:\[V = 2\pi \cdot \frac{2\sqrt{3}}{5} \left( (2)^{5/2} - (0)^{5/2} \right) = 2\pi \cdot \frac{2\sqrt{3}}{5} \cdot 2^{5/2}.\]

Step 6: Final Calculation

Compute \( 2^{5/2} = 4\sqrt{2} \). Therefore,\[V = 2\pi \cdot \frac{2\sqrt{3}}{5} \cdot 4\sqrt{2} = \frac{16\pi \sqrt{6}}{5}.\] This is the volume of the solid.

Key Concepts

Volume of Solid of Revolution

When we talk about the volume of a solid of revolution, we are referring to a method used in calculus to find the three-dimensional volume of shapes created by rotating a two-dimensional area around an axis. This technique is particularly useful because it allows us to evaluate volumes using integration.

Here's how it works:

• A given function or set of lines defines the boundary of a region on a plane.
• We rotate this entire region around a specified axis, which could be the x-axis, y-axis, or any other line.
• The result is a three-dimensional solid whose volume we want to determine.

The overall goal is to set up an integral that, when evaluated, gives us this volume. By understanding the geometry and the rotation involved, we align the integral with the function, taking into account the limits of integration, to solve for the volume.

The method of cylindrical shells, as in our exercise, provides a systematic approach to finalize this calculation.

Cylindrical Shells Method

The cylindrical shells method is one of the techniques used for finding the volume of solids of revolution. It's especially handy when the rotation involves the y-axis or when working with vertical or horizontal strips proves tricky. This method simplifies the process by employing cylindrical elements.

The formula at the heart of the cylindrical shells method is:\[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx, \]where:

• \( x \)
  is the radius of the shell from the axis of rotation.
• \( f(x) \)
  is the height of the shell, given by the function.
• The integral is evaluated between limits \( a \) and \( b \), which are the bounds of the region in question.

This method calculates the volume by summing up the volumes of infinitesimally thin cylindrical shells. Each shell's surface area contributes to the total volume, and integration across the whole range cumulatively finds the overall volume.

Definite Integral

Definite integrals play a crucial role in estimating areas, volumes, and other cumulative quantities in calculus. A definite integral is evaluated over a specific interval, offering an exact numerical result rather than a general formula.
The notation for a definite integral is:\[ \int_{a}^{b} f(x) \, dx, \]where:

• \( a \) and \( b \)
  are the lower and upper bounds, respectively.
• \( f(x) \)
  is the continuous function being integrated.
• The process yields the accumulated sum of all infinitesimal areas under the curve \( f(x) \) from \( x = a \) to \( x = b \).

Definite integrals are essential in determining the volume of solids, as they allow us to add up the infinite slices or shells making up the volume. In our specific problem, we used a definite integral to sum up the cylindrical shells from the region bounded by given functions, finding the solid's total volume.