Problem 49 - Calculus for Biology and Medicine

Question

Find the volumes of the solids obtained by rotating the region bounded by the given curves about the \(x\) -axis. In each case, sketch the region together with a typical disk element. \(y=e^{x}, y=e^{-x}, 0 \leq x \leq 2\)

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{\pi}{2} (e^{4} + e^{-4})\).

Step 1: Understand the Problem

We need to find the volume of the solid formed by rotating the region bounded by the curves \(y = e^{x}\) and \(y = e^{-x}\) about the \(x\)-axis from \(x = 0\) to \(x = 2\). This requires sketching the curves to visualize the region and understand its boundaries.

Step 2: Sketch the Region

Sketch the graphs of \(y = e^{x}\) and \(y = e^{-x}\) on the same coordinate plane. The curve \(y = e^{x}\) increases rapidly, whereas \(y = e^{-x}\) decreases as \(x\) increases. The bounded region is between these curves from \(x = 0\) to \(x = 2\). Label the intersection points at \(x = 0\) and \(x = 2\) to frame the bounded area.

Step 3: Set Up the Volume Integral

Using the disk method, the volume \(V\) of the solid of revolution is given by the integral: \[ V = \pi \int_{0}^{2} (R^2 - r^2) \, dx \]where \(R = e^x\) is the outer radius and \(r = e^{-x}\) is the inner radius of a typical disk.

Step 4: Evaluate the Integral

Substitute the expressions for the radii into the integral:\[ V = \pi \int_{0}^{2} ((e^{x})^2 - (e^{-x})^2) \, dx = \pi \int_{0}^{2} (e^{2x} - e^{-2x}) \, dx \]Evaluate this integral:\[ \int e^{2x} \, dx = \frac{1}{2} e^{2x} + C \]\[ \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} + C \]Plug these into the volume integral:\[ \pi \left[ \frac{1}{2}e^{2x} + \frac{1}{2}e^{-2x} \right]_{0}^{2} \]\[ = \pi \left( \frac{1}{2}e^{4} - \frac{1}{2}e^{-4} - ( \frac{1}{2}e^{0} - \frac{1}{2}e^{0} ) \right) \]Simplify the expression to find the volume.

Step 5: Simplify and Compute

Continuing from the previous step:\[ \pi \left[ \frac{1}{2} (e^{4} + e^{-4}) - 0 \right] \]which simplifies to:\[ \pi \left( \frac{1}{2} e^{4} + \frac{1}{2} e^{-4} \right) = \frac{\pi}{2} (e^{4} + e^{-4}) \]Calculate this expression to get the final volume of the solid.

Key Concepts

Disk Method

When calculating the volume of a solid of revolution, a common approach is the disk method. This technique is particularly useful when you revolve a region around an axis and the resulting solid can be visualized as a series of disk-shaped slices. To apply the disk method, imagine slicing the solid perpendicular to the axis of rotation, in this case, the x-axis. Each slice resembles a disk (or a washer, if there's a hole in the middle), and the thickness of each slice is a small change in the x-direction, typically denoted as \(dx\).

• The radius of each disk is determined by the function that forms the outer boundary of the region. For example, in our problem, the outer function is \(y = e^x\), making the radius of a disk \(R = e^x\).
• If there is a hole (inner boundary), you subtract the inner function from the outer to find the radius. This is the case with the inner function \(y = e^{-x}\), giving an inner radius \(r = e^{-x}\).

By integrating these radii in respect to \(x\), with each slice contributing to the overall volume, the disk method allows you to sum up the volumes of an infinite number of infinitely thin disks.

Solid of Revolution

A solid of revolution refers to a three-dimensional shape created by rotating a two-dimensional region around a line, known as the axis of rotation. In the example given, we rotate the region defined by the curves \(y = e^x\) and \(y = e^{-x}\) about the x-axis, creating a smooth, doughnut-like structure. This process essentially 'sweeps' the planar region through space, forming a symmetrical shape around the axis. Understanding the following will help:

• **Axis of Rotation:** Here it is the x-axis, meaning each point on the boundary of the region rotates horizontally.
• **Boundaries of the Region:** These are defined by the curve equations, starting from \(x = 0\) to \(x = 2\).
• **Intersection Points:** These occur where the functions meet, framing the region you need to rotate. Identifying these correctly forms an accurate visual model of the solid.

By rotating the region around the x-axis, we effectively create a three-dimensional volume that can be calculated using integral calculus.

Integral Calculus

Integral calculus is the branch of mathematics that deals with the accumulation of quantities, such as areas under curves or volumes of solids. To find the volume of a solid of revolution, integral calculus provides the tools to add up the infinite slices or disks you'll conceptualize.In our problem, the integral calculus is set up as the integral of the difference between the outer and inner function squared:\[ V = \pi \int_{a}^{b} (R^2 - r^2) \, dx \]Where:

• \(R = e^x\), representing the outer radius function.
• \(r = e^{-x}\), representing the inner radius function.
• \([a, b] = [0, 2]\), representing the bounds of our integration, i.e., the section of the x-axis being used. Using the properties of integrals, each evaluated piece, \(\int e^{2x} \,dx\) and \(\int e^{-2x} \,dx\), finds the area under the curve, effectively building up the complete volume of the solid.

By setting up and evaluating these integrals, we can find the exact volume of the solid formed, showcasing the powerful precision of integral calculus.