Question

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(x\) -axis. $$y=x, y=2-x, \text { and } y=0$$

Short answer

Answer: The volume of the solid generated when region R is revolved around the x-axis is V = \(\frac{15\pi}{32}\).

Step-by-step solution

Visualize and understand the region R

First, we need to understand the region R that is bounded by y=e^(-x), y=0, x=0, and x= ln(4). We can imagine that on the xy-plane, we have the exponential curve y=e^(-x) starting from the point (0,1) and approaching the x-axis as x goes to infinity. The region R is a right triangle, with vertices at (0,0), (0,1), and (ln(4), 1/4).

Set up the integral expression using the disk method

To find the volume of the solid generated when region R is revolved about the x-axis, we will use the formula for the volume of a solid using the disk method: $$V = \pi \int_a^b [f(x)]^2 dx$$ where V is the volume, a and b are the limits of integration on the x-axis, and f(x) is the function representing the curve. In our case, the function f(x) is given by f(x) = e^(-x), and the interval of integration is [0, ln(4)]. So, the volume can be expressed as: $$V = \pi \int_0^{\ln 4} (e^{-x})^2 dx$$

Evaluate the integral to find the volume

Now, we need to evaluate the integral to find the volume: $$V = \pi \int_0^{\ln 4} e^{-2x} dx$$ To evaluate this integral, we can use substitution method. Let u=-2x, then du=-2 dx. So, dx = (-1/2) du. Now the integral becomes: $$V = \pi \int_{0}^{-2\ln 4} (-\frac{1}{2}) e^u du$$ We can take the constant out of the integral: $$V = -\frac{\pi}{2} \int_{0}^{-2\ln 4} e^u du$$ Now, integrate e^u with respect to u: $$V = -\frac{\pi}{2} [e^u]_{0}^{-2\ln 4}$$ $$V = -\frac{\pi}{2} (e^{-2\ln 4} - e^0)$$ Use the property of exponentials: e^(ln(a^b))= a^b, in our case, e^(-2ln 4) = (1/4^2) = 1/16. Therefore, $$V = -\frac{\pi}{2} (\frac{1}{16} - 1)$$ $$V = \frac{\pi}{2} (1 - \frac{1}{16})$$ $$V = \frac{15\pi}{32}$$ Hence, the volume of the solid generated when region R is revolved about the x-axis is V = \(\frac{15\pi}{32}\).

Key concepts

Disk Method

The Disk Method is a technique used to calculate the volume of a solid of revolution created by rotating a region around an axis. It's particularly useful when dealing with solids whose cross-sections are circular disks. The concept is simple: imagine slicing the solid into thin circular disks perpendicular to the axis of rotation.

Here's how it works:

• Identify the region: You need a region that can be revolved around an axis. In our example, this region is bounded by the curve, horizontal lines, and vertical lines.
• Define the function: The boundary of the region is represented by a function, in this case, the exponential function, such as \(y = e^{-x}\).
• Determine the limits: These are usually x-values that mark the start and end of the region on the axis of rotation. For us, it's from \(x = 0\) to \(x = \ln(4)\).
• Apply the formula: The disk method formula is \[ V = \pi \int_a^b [f(x)]^2 \, dx \] where \(f(x)\) is the radius of each disk, and \(a\) and \(b\) are the integration limits.

Definite Integral

A Definite Integral is a mathematical tool used to find the area under a curve within specific limits. When using the disk method to find volume, you compute a definite integral over the interval defined by the bounds of the region.

The steps in our calculation are as follows:

• Setting up the integral: For this problem, we set up an integral to calculate the volume, given by \( V = \pi \int_0^{\ln 4} (e^{-x})^2 \, dx \).
• Compute the integral: This step often involves basic calculus techniques such as substitution or integration by parts. Here, we used substitution to simplify the integration process.
• Evaluate the integral: Once the integral is solved, plug in the upper and lower bounds, subtract, and multiply by any constants from the disk method formula.

Definite integrals ensure that you have a detailed solution that gives the exact volume enclosed by the surface and the axis, rather than merely a general form.

Exponential Functions

Exponential Functions are mathematical functions representing an expression of the form \( y = a^x \). These functions are crucial in modeling real-world phenomena like population growth, radioactive decay, and more. In calculus, they often appear in growth or decay models.

In this problem, the function we're focusing on is \( y = e^{-x} \), where:

• Base \(e\): This is the natural exponential base, ≈ 2.718. It appears widely across calculus due to unique properties, such as the derivative and integral of \( e^x \) being \( e^x \).
• Negative Exponent: Functions like \( e^{-x} \) decay rapidly, creating curves that head toward zero as \(x\) increases.
• Application to Volume: The key part of using exponential functions here is that each point along \( e^{-x} \) represents a different radius for the disks produced by rotation. Thus, \( (e^{-x})^2 \) forms the basis of the volume integral using the disk method.

Exponential functions provide the curves that define the boundary of the region we're rotating, showing how mathematics can turn abstract functions into concrete shapes.