Problem 22 - Essential Calculus

Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(x\) -axis. $$ y=e^{x / 2}, \quad y=0, \quad x=0, \quad x=4 $$

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the region bounded by the graphs of the given equations about the \(x\)-axis is \(\pi[e^4 - 1]\) cubic units.

Step 1: Identify the Bounded Region

The region bounded by the graphs of the equations \(y=e^{x / 2}\), \(y=0\), \(x=0\), and \(x=4\) should be identified. It is bounded below by the \(x\)-axis (\(y=0\)), on the left by the \(y\)-axis (\(x=0\)), on the right by the vertical line (\(x=4\)), and above by the graph of the equation \(y=e^{x / 2}\).

Step 2: Implement the Disk Method

Use the Disk Method to solve this. It involves integrating the cross-sectional area of the solid. For disks/washers, the trivial cross section is a circle, with an area \(A = \pi r^2\). Thus, the differential volume element \(dV = A\: dx = \pi [r(x)]^2\: dx\). Here, \(r(x) = y = e^{x / 2}\), since the radius of a disk here will be the distance from the \(x\)-axis to a point on the curve \(y=e^{x / 2}\). So, \(dV = \pi [e^{x / 2}]^2\: dx\). Before substituting this into volume integral, simplify \(r(x)\) to work with.

Step 3: Simplify and Setup the Volume Integral

\([r(x)]^2 = [e^{x / 2}]^2 = e^x\). Substitute this into the volume integral and set the limits of integration from \(x=0\) to \(x=4\). This gives: \(V = \int_{0}^{4} \pi e^x\: dx\).

Step 4: Evaluate the Definite Integral

Evaluate the definite integral \(V = \int_{0}^{4} \pi e^x\: dx = \pi[e^x]_{0}^{4} = \pi[e^4 - e^0] = \pi[e^4 - 1]\).

Key Concepts

Disk Method

The Disk Method is a technique for finding the volume of a solid of revolution. Imagine taking a region on a plane and revolving it around an axis to create a three-dimensional shape. By slicing this shape perpendicular to the axis of revolution, you form disks or washers. The volume can be found by summing up the volumes of these infinitesimally thin disks.

To apply it:

• Identify the function that describes the shape's boundary.
• Determine the radius of the disk, which is the distance from the axis of rotation to the curve.
• Set up the integral that sums the disk's areas over the given interval. The formula for volume is:

The volume of revolution using this method is expressed as:\[V = \int_{a}^{b} \pi [f(x)]^2 \, dx\]
For example, if revolving around the x-axis, the radius is simply the function value, \(f(x)\). This method is straightforward when the function is easily integrated.

Definite Integral

The concept of definite integrals is fundamental in calculus and is used to compute quantities like areas under curves and volumes. It represents the net area between the graph of a function and the x-axis within a specified interval.

When dealing with solids of revolution, the definite integral helps determine the exact volume by integrating the cross-sectional areas (like disks). For the exercise involving the function \(y = e^{x/2}\):

• The region of interest is from \(x=0\) to \(x=4\).
• We integrate the square of the function because we consider the area of a circle, which is \(\pi r^2\).
• This gives the integral \(\int_{0}^{4} \pi e^x \, dx\), where \(e^x\) is the simplified form of \([e^{x/2}]^2\).

Evaluating a definite integral involves finding the antiderivative and calculating the difference between its values at the upper and lower limits of integration. Ultimately, this integral tells us the total space the solid occupies.

Exponential Functions

Exponential functions are a type of mathematical function where a constant base is raised to a variable exponent. In calculus, they often appear in problems involving growth and decay, and in our exercise, play a vital role in defining the curve to be revolved.

The function we are examining is \(y = e^{x/2}\), which is an exponential function with the base \(e\), a mathematical constant approximately equal to 2.718. This specific function describes a curve that increases rapidly as \(x\) increases, indicating exponential growth.

• The nature of exponential functions means that as \(x\) grows, the values of \(y\) increase multiplicatively, not just additively.
• These functions are particularly smooth and continuous, making them well-suited for integration.
• The exponential function's derivative is notably equal to itself, \( \frac{d}{dx} e^u = e^u \frac{du}{dx} \), which simplifies calculus operations.

Understanding exponential functions can improve comprehension of real-world phenomena like population growth or radioactive decay, and in this case, helps in precisely evaluating the volume of the solid.