Problem 46 - Calculus of a Single Variable

Question

Analyzing an Integral In Exercises \(43-46,\) the integral represents the volume of a solid of revolution. Identify (a) the plane region that is revolved and (b) the axis of revolution. $$ 2 \pi \int_{0}^{1}(4-x) e^{x} d x $$

Step-by-Step Solution

Answer

a) The plane region that is revolved is under the curve \(y=(4-x)e^{x}\) from x=0 to x=1. b) The axis of revolution is the x-axis.

Step 1: Identify The Revolved Region

The definite integral is given by \(2 \pi \int_{0}^{1}(4-x)e^{x} dx\). The integrand function is \((4-x)e^{x}\) and spans from 0 to 1 which suggests that the revolved region is the area under the curve \(y=(4-x)e^{x}\) from the interval x=0 to x=1.

Step 2: Identify The Axis Of Revolution

Since there is no value of x in the brackets that's subtracted from the function, we can identify that the axis of revolution is the x-axis.

Step 3: Visual Representation

To better understand, one can sketch the curve \(y=(4-x)e^{x}\)\, on the interval [0,1] and revolve it around the x-axis to visualize the solid of revolution we're working with. Doing so, we can see that our solid is a kind of truncated cone shape, tapering off as x approaches 1.

Key Concepts

Definite Integral

In calculus, the definite integral is an essential concept that helps in finding the accumulated sum of quantities that can vary. Think of it like a high-precision way to add up a bunch of tiny values over a specific interval.

For example, when you calculate the definite integral of a function between two points on the x-axis, you're essentially summing up the area under the curve from one point to another. Here, the symbols \( \int_{a}^{b} \) denote the integral from point \(a\) to point \(b\).

In the context of the given exercise, the definite integral \(2 \pi \int_{0}^{1}(4-x) e^{x} dx\) is used to calculate the volume of a solid of revolution. This process involves 'spinning' the area under the curve around an axis to get a 3D shape. The \(2 \pi\) factor suggests we are dealing with circular symmetry, as we often do when calculating volumes of solids of revolution.

Area Under the Curve

The area under the curve of a function on a graph represents all the points between the function's curve and the x-axis, within a specified interval.

This area can be visualized as a slice of land between a road (the curve) and the flat ground (the x-axis) stretching between two fences (the points \(a\) and \(b\) on the x-axis). To find this area, mathematicians use integration. The larger the area, the greater the integral's value will be.

In our example, the function \(y=(4-x)e^{x}\) defines the curve. To find the area under it from \(x=0\) to \(x=1\), one would calculate the definite integral. It's like adding up all the 'slices' of areas you get when you take vertical slices of the graph, then multiply them by their width, making them infinitesimally thin, and adding them all together.

Axis of Revolution

The axis of revolution is the line around which a two-dimensional shape rotates to form a three-dimensional object. Think of it like the central pole of a merry-go-round; everything revolves around it.

In our exercise, by identifying that the area under the curve has no horizontal shift (since there's no subtraction of \(x\) from the function in the brackets), we infer that the curve rotates around the x-axis. This is our axis of revolution.

Imagine the 2D area under the curve of \(y=(4-x)e^{x}\) as a flat sheet of paper. When we 'revolve' or spin this paper around the x-axis, we transform it into a 3D shape, such as a vase or a bell, based on the shape of the curve. If the axis of revolution changes, say to the y-axis or any other line, the resulting solid's shape would change dramatically.