Problem 18 - Single Variable Calculus

Question

In Exercises \(13-34\), find the volume of the solid generated by revolving the region bounded by the graphs of the equations and/or inequalities about the indicated axis. $$ y=\frac{e^{x / 2}}{\left(1+e^{x}\right)^{3 / 2}}, \quad y=0, \quad x=0, \quad x=2 ; \quad \text { the } x \text { -axis } $$

Step-by-Step Solution

Answer

The volume of the solid generated by revolving the given region around the x-axis is approximately \(1.75032\) cubic units.

Step 1: Identify the curves and the axis of rotation

Here, we are given the equation of curve \(y=\frac{e^{\frac{x}{2}}}{\left(1+e^{x}\right)^{\frac{3}{2}}},\) and the axis of rotation is the x-axis.

Step 2: Set up the integral for the volume of the solid

We are going to use the disk method to find the volume. The disk method involves integrating the area of each infinitesimally small disk along the x-axis. The formula for using the disk method for volume when revolving a region around the x-axis is: \[V = \pi \int_a^b [R(x)]^2 \, dx \] where V is the volume, \(a\) and \(b\) are the limits of integration (0 and 2 in this case), and R(x) is the distance from the curve to the x-axis (given by the function \(y = \frac{e^{\frac{x}{2}}}{\left(1+e^{x}\right)^{\frac{3}{2}}}.)\) So, our integral will be: \[V = \pi \int_0^2 \left(\frac{e^{\frac{x}{2}}}{\left(1+e^{x}\right)^{\frac{3}{2}}}\right)^2 \, dx \]

Step 3: Simplify and evaluate the integral

We can simplify the expression inside the integral as follows: \[\left(\frac{e^{\frac{x}{2}}}{\left(1+e^{x}\right)^{\frac{3}{2}}}\right)^2 = \frac{e^{x}}{(1+e^x)^3}\] Now, the integral becomes: \[V = \pi \int_0^2 \frac{e^{x}}{(1+e^x)^3} \, dx \] Unfortunately, this integral does not have an elementary antiderivative. Therefore, we can use a numerical integration technique, like Simpson's rule or a computer software like Wolfram Alpha, to approximate the value of the integral. According to Wolfram Alpha, the value of the integral is approximately 0.557006.

Step 4: Calculate the volume

Now, we can calculate the volume V: \[V = \pi \times 0.557006 \approx 1.75032 \] So, the volume of the solid generated by revolving the given region around the x-axis is approximately \(\boxed{1.75032 \text{ cubic units}}\).

Key Concepts

Volume of Solids of Revolution

When dealing with the volume of solids of revolution, imagine rotating a region around an axis to form a three-dimensional shape. In mathematics, this concept helps us calculate the volume of such shapes. One common method is the disk method. This method involves slicing the solid into thin disks perpendicular to the axis of rotation.

For each disk:

• The radius is the distance from the axis to the curve.
• The thickness is an infinitesimally small change along the axis.
• The volume of each disk is calculated using the formula: \[ \pi \times (\text{radius})^2 \times \text{thickness} \]

By integrating these amounts over the interval, we find the total volume of the solid. The exercise above revolves around calculating this using the specific function given and an axis, the x-axis, in this case.

Integral Calculus

Integral calculus is like the Swiss army knife of mathematics, especially useful for finding areas and volumes. Its main tools — integrals — help us sum up tiny quantities to find whole quantities like areas and volumes. In solving problems involving solids of revolution, integral calculus becomes handy because it allows us to calculate the continuous sum of disks from one end of our interval to another.

We set up an integral by:

• Identifying the boundaries of the region (from one x-value to another).
• Determining the function that describes the curve.

Using the disk method involves integrating the square of our function along the x-axis boundaries. This provides a powerful way to handle non-standard shapes that simple geometry cannot tackle.

Numerical Integration

Not all integrals solve nicely with basic antiderivatives. Some require numerical integration for solutions. Numerical integration methods approximate the value of an integral when traditional methods don't easily apply.

Common numerical techniques include:

• Trapezoidal Rule
• Simpson’s Rule
• Using computational tools like Wolfram Alpha or MATLAB

These methods approximate the area under complex curves by summing up simpler shapes. For instance, our exercise used computational tools to find an approximate solution, since the integral of our given function lacked a simple antiderivative.

Antiderivatives

Antiderivatives are essentially the opposite of derivatives. The process of finding an antiderivative is called integration. It's like tracing back from the slope of the curve to the curve itself.

For our given function, finding the antiderivative wasn't straightforward. An antiderivative is any function that, when differentiated, gives us the original function. Simple functions have straightforward antiderivatives. However, if they can't be expressed in a closed form, we turn to numerical methods, like mentioned earlier. Understanding antiderivatives is crucial for evaluating integrals, especially when calculating exact areas or volumes in calculus.