Problem 15 - University Calculus: Early Transcendentals

Question

Write an iterated integral for \(\iint_{R} d A\) over the described region \(R\) using (a) vertical cross-sections, (b) horizontal cross-sections. Bounded by \(y=e^{-x}, y=1,\) and \(x=\ln 3\)

Step-by-Step Solution

Answer

(a) \(\int_{0}^{\ln 3} \int_{e^{-x}}^{1} dy \, dx\); (b) \(\int_{0}^{1} \int_{0}^{-\ln y} dx \, dy\).

Step 1: Understanding the Region

The region \(R\) is bounded by the curves \(y=e^{-x}\), \(y=1\), and the line \(x=\ln 3\). The boundaries imply a region in the xy-plane where \(x\) varies from 0 to \(\ln 3\) and within this interval, \(y\) varies between the curve \(y=e^{-x}\) and the line \(y=1\).

Step 1: Setup for Vertical Cross-Sections

To use vertical cross-sections, we keep \(x\) constant and integrate with respect to \(y\). The inner integral defines the bounds for \(y\): it goes from \(y=e^{-x}\) to \(y=1\). The outer integral then integrates these results with respect to \(x\) over the interval from 0 to \(\ln 3\). The iterated integral becomes: \[ \int_{0}^{\ln 3} \int_{e^{-x}}^{1} dy \, dx \]

Step 2: Setup for Horizontal Cross-Sections

To use horizontal cross-sections, we keep \(y\) constant and integrate with respect to \(x\). Here, \(x\) bounds depend on \(y\): \(x\) ranges from \(x = 0\) to \(x = -\ln y\) since solving \(y = e^{-x}\) for \(x\) gives \(x = -\ln y\). The outer integral integrates these results with respect to \(y\) from the lower bound \(y=e^{-x}, x=0\), which gives \(y=1\), and \(y\) itself to 1. The iterated integral is: \[ \int_{0}^{1} \int_{0}^{-\ln y} dx \, dy \]

Step 4: Verify Boundaries and Setup

To double-check, confirm the region intersects at \((0, 1)\) since \(y = e^{-0} = 1\) and confirms within bounds at the upper limits as well. Ensure both integrals logically represent the entire area \(R\), addressing possible confusion from different approaches.

Key Concepts

Vertical Cross-Sections

When solving an integral problem using vertical cross-sections, you imagine slicing your region in the direction parallel to the y-axis. This means you hold the x-value constant while integrating over changing y-values.
In the original exercise, the region of interest is bounded by the curve \(y = e^{-x}\), the line \(y = 1\), and \(x = \ln 3\). To describe the region using vertical cross-sections:
- Let \(x\) vary from 0 to \(\ln 3\). This defines the vertical range of slices.- For each slice, \(y\) varies from \(y = e^{-x}\) to \(y = 1\). This determines the integral with respect to y.The iterated integral takes the form \[ \int_{0}^{\ln 3} \int_{e^{-x}}^{1} dy \, dx \] This setup considers slices that sweep horizontally across your defined range for \(x\), collecting the area under the defined curves and lines. Each vertical slice is a small rectangular area, and integrating over \(x\) compiles them into one total area.

Horizontal Cross-Sections

Horizontal cross-sections provide a flipped perspective on integration. Here, you hold y constant and integrate along the x-axis, parallel to it, observing how y-values range for specific x-values.
Given the same region bounded by \(y = e^{-x}\), \(y = 1\), and \(x = \ln 3\), this approach analyzes how the x-values change within the bounds of constant y-slices.
- Fix \(y\) and observe that the right and left boundaries for \(x\) are 0 and \(-\ln y\), respectively (since rearranging \(y = e^{-x}\) gives \(x = -\ln y\)).- The outer integral will consider the range for \(y\) from 0 to 1.
Hence, the iterated integral appears as:\[ \int_{0}^{1} \int_{0}^{-\ln y} dx \, dy \]This method reflects slices that are horizontally layered, each representing a line's portion without altering y. Every horizontal slice adds potency to the total area by gauging the span along the x-axis for every fixed \(y\).

Bounded Regions in Calculus

In calculus, bounded regions help us visualize and measure spaces confined by curves and lines. Think of them as shapes whose borders are determined by mathematical relations. Understanding boundaries is vital for calculating areas using integrals.
For the problem at hand:- The boundary curves and lines are \(y = e^{-x}\), \(y = 1\), and \(x = \ln 3\).- These lines define what points belong to the interior region \(R\) which is the subject of our integration.
It's crucial to picture the overlap and differences between the boundaries. This improves comprehension about how the integrals leap through infinitesimal slices, whether vertically or horizontally, to cover the entire specified region. By visualizing boundaries:- You ensure the limits in your integral are accurate.- Each approach (vertical or horizontal) precisely encloses the area inside the perimeter.Using iterated integrals over bounded regions invites efficient computations of areas in multi-variable calculus, unlocking insights beyond straightforward single-variable scenarios.