Problem 49 - Calculus Early Transcendentals

Question

Find the volume of the solid of revolution. Sketch the region in question. The region bounded by \(y=e^{x}, y=0, x=0,\) and \(x=2\) revolved about the \(x\) -axis

Step-by-Step Solution

Answer

Question: Find the volume of the solid generated by revolving the region bounded by the curves \(y=e^x\), \(y=0\), \(x=0\), and \(x=2\) around the x-axis. Answer: The volume of the solid of revolution is \(\frac{\pi}{2}(e^4 - 1)\).

Step 1: Sketch the region

First, we need to sketch the region bounded by the given curves. For this, draw the curve of \(y=e^x\), the lines \(y=0\), \(x=0\), and \(x=2\) on the xy-plane. The region enclosed by these curves will be revolved around the x-axis to generate the solid of revolution.

Step 2: Set up the integral for the volume

Using the disk method, the volume of the solid of revolution can be found using the integral formula: $$ V = \pi\int_a^b [f(x)]^2 dx $$ In this case, \(f(x) = e^x\), \(a=0\), and \(b=2\). So, the integral becomes: $$ V = \pi\int_0^2 (e^x)^2 dx $$

Step 3: Solve the integral

Simplify the integral: $$ V = \pi\int_0^2 e^{2x} dx $$ Now, integrate \(e^{2x}\) with respect to \(x\): $$ V = \pi\left[\frac{1}{2}e^{2x}\right]_0^2 $$

Step 4: Evaluate the limits

Plug in the limits of integration into the integral expression: $$ V = \pi\left[\frac{1}{2}e^{2(2)} - \frac{1}{2}e^{2(0)}\right] $$

Step 5: Simplify the expression and find the volume

Simplify and compute the result: $$ V = \pi\left[\frac{1}{2}e^4 - \frac{1}{2}\right] $$ The volume of the solid of revolution is: $$ V = \frac{\pi}{2}(e^4 - 1) $$

Key Concepts

Disk Method

When you are trying to find the volume of a solid of revolution, the Disk Method is a very useful tool. It is particularly applicable when a region is revolved around an axis, like the x-axis, to create a solid. Think of the shape being made up of many thin disk-like slices stacked along that axis. Each disk has a small thickness of \(dx\) and a radius that corresponds to the function value at that point.

Here's how you can use the Disk Method to calculate the volume of a solid:

• Identify the region that will be revolved — in our exercise, it's an area bounded by \(y = e^{x}\), the x-axis, and the vertical lines \(x = 0\) and \(x = 2\).
• Use the disk formula \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx\), where \(f(x)\) represents the height of the disk and \(a\) and \(b\) are the bounds of integration.
• Here, the function \(f(x) = e^{x}\), and you need to integrate from \(x = 0\) to \(x = 2\).

Remember, each slice or disk has a thickness \(dx\), and the radius of each disk is simply \(f(x)\). Multiply the area of the disk \((\pi [f(x)]^2)\) by \(dx\) to get the volume of each slice, and sum this using an integral to find the entire volume.

Definite Integral

The definite integral is a crucial concept in calculus, especially when calculating areas under a curve or volumes of shapes. A definite integral gives you the accumulated quantity, in this case, volume, from one point to another on the x-axis.

To compute the volume of a solid using integration, perform these steps:

• Set the bounds of integration. In the given example, these bounds are \(a = 0\) and \(b = 2\), as this is the interval over which the region is revolved around the x-axis.
• Integrate the function squared, \(f(x)^2\), over these bounds. Here, that is integrating \(e^{2x}\) from \(x = 0\) to \(x = 2\).
• Use the Fundamental Theorem of Calculus to evaluate this integral. Essentially, calculate the antiderivative of the function and then substitute the upper and lower bounds into this antiderivative.

For our problem, this results in the expression \(\pi \left[ \frac{1}{2}e^{2x} \right]_{0}^{2}\). Subtracting the antiderivative value at the lower bound from the value at the upper bound will give you the total volume.

Exponential Function

Exponential functions, like \(y = e^{x}\), are fundamental in mathematics due to their unique growth properties. For the exercise at hand, \(e^{x}\) not only defines the curve we revolve but also affects the volume of the solid of revolution.

Understanding exponential functions is critical for solving such volume problems:

• They have the form \(y = e^{x}\), where \(e\) is Euler's number, approximately 2.718.
• These functions grow rapidly, meaning the volume they enclose changes significantly as \(x\) increases.
• When you square an exponential function, such as in our volume formula \(e^{2x}\), it grows even faster.

In the exercise, integrating \(e^{2x}\) helps find the volume under the curve from \(x = 0\) to \(x = 2\), providing the exponential growth component necessary to derive the correct volume value. Calculating \(\frac{1}{2}e^{2x}\) in this kind of integral is standard, ensuring the exponential effects are captured accurately.