Problem 5 - Calculus Single Variable

Question

In each of Exercises 1-6, use the method of disks to calculate the volume \(V\) of the solid that is obtained by rotating the given planar region \(\mathcal{R}\) about the \(x\) -axis. \(\mathcal{R}\) is the region above the \(x\) -axis, below the graph of \(y=\) \(\exp (x)\), to the right of \(x=0\), and to the left of \(x=\ln (2)\).

Step-by-Step Solution

Answer

The volume of the solid is \( \frac{3\pi}{2} \).

Step 1: Identify the Region

The given region \( \mathcal{R} \) is above the \( x \)-axis, below the graph of \( y = e^x \), to the right of \( x = 0 \), and to the left of \( x = \ln(2) \). This defines a vertical strip bounded by these lines and curves.

Step 2: Set Up Disk Method Formula

The method of disks involves finding the volume of a solid of revolution by integrating the cross-sectional area of disks perpendicular to the axis of rotation. Here, the axis of rotation is the \( x \)-axis. The radius of each disk is the distance from the \( x \)-axis to the curve, \( y = e^x \). The volume \( V \) of the solid is given by the integral \[V = \pi \int_{0}^{\ln(2)} (e^x)^2 \, dx.\]

Step 3: Simplify the Integrand

The integrand \( (e^x)^2 = e^{2x} \). Thus, the integral for the volume becomes \[V = \pi \int_{0}^{\ln(2)} e^{2x} \, dx.\]

Step 4: Integrate the Function

To integrate \( e^{2x} \), use the substitution rule. Let \( u = 2x \) which implies \( du = 2 \, dx \) or \( dx = \frac{1}{2} \, du \). Rewriting the integral, we get \[\frac{\pi}{2} \int e^u \, du.\]After integration, \[\frac{\pi}{2} e^u = \frac{\pi}{2} e^{2x}.\]

Step 5: Evaluate the Definite Integral

Now, substitute back and use the limits of integration from 0 to \( \ln(2) \):\[V = \frac{\pi}{2} [e^{2 \times \ln(2)} - e^{2 \times 0}].\]Since \( e^{2 \times \ln(2)} = e^{\ln(4)} = 4 \) and \( e^0 = 1 \), we have\[V = \frac{\pi}{2} (4 - 1) = \frac{3\pi}{2}.\]

Step 6: Conclusion

The volume of the solid obtained by rotating the region \( \mathcal{R} \) about the \( x \)-axis is \( \frac{3\pi}{2} \).

Key Concepts

Disk Method

The Disk Method is a way to find the volume of a solid of revolution. When you rotate a region about an axis, this method allows us to calculate the volume of the resulting 3D shape. Imagine cutting the solid into thin slices, like slicing a loaf of bread, perpendicular to the axis of rotation. Each slice is a disk.

• The axis of rotation in our example is the x-axis.
• The formula involves integrating the area of each disk across the region.

The radius of each disk is the distance from the axis to the curve. If our region is bounded by the function \( y = e^x \), the radius is simply \( e^x \). The area of a disk is the area of a circle, \( \pi \times \text{radius}^2 \). So, here, it's \( \pi \times (e^x)^2 \). By integrating this expression with respect to x, we find the total volume.

Integration

Integration is a mathematical process used to find areas, volumes, and many other physical properties. It is essentially the reverse operation to differentiation. When we talk about finding the volume of a solid of revolution using the disk method, integration helps us sum up all the tiny slices.

Just as differentiation tells us the rate of change, integration gives us a way to accumulate quantities. For example, in our volume calculation, we're using integration to add up the areas of many disks:

• First, we set up the disk's area in terms of variables.
• Then, we integrate across the bounds of the region.

This allows us to find the entire volume. Using the definite integral from the region's bounds, here from 0 to \( \ln(2) \), ensures our answer represents the real-world volume.

Definite Integral

When we use integration to find exact values over a specific interval, we use what's called a definite integral. In contrast to an indefinite integral, which represents a general form of a function's antiderivative, a definite integral gives a numerical result.

For our exercise, the definite integral captures the complete volume of the solid from one specific boundary to another:

• The lower bound \( x = 0 \) and the upper bound \( x = \ln(2) \) define the strip of the region being revolved.
• The function \( e^{2x} \) within the integral \( \pi \int_{0}^{ ext{ln}(2)} e^{2x} \, dx \) represents the squared radius of each disk as it's rotated.

When we evaluate this integral, we obtain the volume \( \frac{3\pi}{2} \), the total space the solid occupies.

Exponential Function

An exponential function is a mathematical function of the form \( y = e^x \), where \( e \) is the base of natural logarithms, approximately 2.71828. Such functions grow rapidly and are notable for their unique properties in calculus. One thing that makes them special is how they transform under operations like differentiation and integration.

In our exercise, \( y = e^x \) defines the top boundary of the region we are revolving:

• The position of this curve changes across the x-axis, defining different radii for our disks.
• As we integrate \((e^x)^2\), or \(e^{2x}\), it showcases how exponential functions handle power transformations.

Understanding exponential growth is key in analyzing many physical phenomena and scenarios where this type of rapidly increasing function appears.