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Chapter 4: Problem 28

Let \(D\) be the region bounded by \(y=1, y=x, y=\ln x\), and the \(x\) -axis. a. Show that \(\iint_{D} y d A=\int_{0}^{1} \int_{0}^{x} y d y d x+\int_{1}^{e} \int_{\ln x}^{1} y d y d x\) by dividing \(D\) into two regions of Type I. b. Evaluate the integral \(\iint_{D} y d A\).

Short Answer

Expert verified
The integral evaluates to \(\frac{3e-4}{6}\).

Step by step solution

01

Describe the Region D

Region D is bounded by the curves: \(y = 1\), \(y = x\), \(y = \ln x\), and the \(x\)-axis. The points of intersection between these curves are found by solving the equations pairwise. Solving \(y = x\) with \(y = \ln x\), we find the intersection occurs at \(x = 1\). The other intersections occur at \(x = e\) for \(y = 1\) and \(y = \ln x\). The region is also bounded on the left by \(x = 0\).
02

Divide D into Two Type I Regions

To divide \(D\) into Type I regions, we consider the vertical lines from a minimum to a maximum x-coordinate, with y-coordinates varying between the two bounds. First, consider Region 1: for \(0 \leq x \leq 1\), \(y\) varies between the \(x\)-axis (\(y = 0\)) and \(y = x\). For Region 2: for \(1 \leq x \leq e\), \(y\) varies between the \(x\)-axis (\(y = \ln x\)) and the line \(y = 1\).
03

Write the Double Integral

The double integral over the entire region \(D\) is split into two integrals corresponding to the two Type I regions. The first integral, representing the area under Region 1, is \(\int_{0}^{1} \int_{0}^{x} y \; dy \, dx\). The second integral, representing Region 2, is \(\int_{1}^{e} \int_{\ln x}^{1} y \; dy \, dx\). These integrals match the description given in part (a) of the problem.
04

Solve the First Integral for Region 1

Evaluate \(\int_{0}^{1} \int_{0}^{x} y \, dy \, dx\). First, integrate with respect to \(y\):\[\int_{0}^{x} y \, dy = \left[ \frac{1}{2}y^2 \right]_0^x = \frac{x^2}{2}.\]Now, integrate the result with respect to \(x\):\[\int_{0}^{1} \frac{x^2}{2} \, dx = \left[ \frac{1}{6}x^3 \right]_0^1 = \frac{1}{6}.\]
05

Solve the Second Integral for Region 2

Evaluate \(\int_{1}^{e} \int_{\ln x}^{1} y \, dy \, dx\). First, integrate with respect to \(y\):\[\int_{\ln x}^{1} y \, dy = \left[ \frac{1}{2}y^2 \right]_{\ln x}^{1} = \frac{1}{2} - \frac{1}{2}(\ln x)^2.\]Now, integrate the result with respect to \(x\): \[\int_{1}^{e} \left(\frac{1}{2} - \frac{1}{2}(\ln x)^2 \right) \, dx.\]Split into two integrals: \(\int_{1}^{e} \frac{1}{2} \, dx - \int_{1}^{e} \frac{1}{2}(\ln x)^2 \, dx\). The first is \([\frac{1}{2}x]_1^e = \frac{e-1}{2}\). Solve the second via substitution \(u = \ln x, du = \frac{1}{x}dx\). Evaluate \(\int_{0}^{1} u^2 \, du = \left[ \frac{u^3}{3} \right]_0^1 = \frac{1}{3}\). Multiply by -1/2: \(-\frac{1}{6}\). Thus, \(\int_{1}^{e} \frac{1}{2}(\ln x)^2 \, dx = \frac{1}{6}\). Combine: \(\frac{e-1}{2} - \frac{1}{6} = \frac{3e-5}{6}\).
06

Combine Results

Sum the results from both regions: \(\frac{1}{6} + \frac{3e-5}{6} = \frac{3e-4}{6}\). This is the value of the double integral over the binary division of region \(D\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Region of Integration
When dealing with double integrals, understanding the "region of integration" is crucial. This refers to the specific area within the coordinate plane where the integration takes place鈥攅ssentially the space we're analyzing. For the region in this problem, denoted as \(D\), it's bound by the curves \(y = 1\), \(y = x\), \(y = \ln x\), and the \(x\)-axis. In Cartesian coordinates, the region looks like a sort of irregular wedge. We determine the boundaries of \(D\) by looking at the intersections of these curves:
  • The line \(y = x\) intersects \(y = \ln x\) at \(x = 1\).
  • \(y = 1\) intersects \(y = \ln x\) at \(x = e\).
  • On the left side, the region is confined by \(x = 0\).
Once these intersections are determined, it becomes easier to visualize where the integral will be applied. Understanding these boundaries is key to setting up the correct limits of integration, which we'll discuss next.
Type I Regions
When slicing a region for double integrals, one common approach is using "Type I regions". These are regions where, for a fixed \(x\), the variable \(y\) ranges between two continuous functions: a lower bound and an upper bound.
For the region \(D\) in this problem, we divide it into two Type I regions:1. **Region 1:** - Here, \(x\) ranges from 0 to 1. - For each \(x\), \(y\) starts at 0 (the x-axis) and goes up to \(y = x\).2. **Region 2:** - In this part, \(x\) spans from 1 to \(e\). - For these \(x\)-values, \(y\) ranges from \(y = \ln x\) to 1.
By dividing the region this way, we make sure that the double integral is more manageable. Type I regions simplify integration by focusing on how one variable (in this case, \(y\)) varies between two functions of another variable. Such ordering makes the calculations straightforward but requires clear understanding of the geometry involved.
Integration Limits
Setting "integration limits" correctly helps determine the scope of our double integration and is essential for accurate calculations.In this exercise, separating the entire region \(D\) into two Type I regions aids in placing suitable integration limits:
  • For **Region 1**, the integration first happens with \(y\): - The inner limits of integration for \(y\) go from 0 to \(x\). - The outer integral for \(x\) goes from 0 to 1.
  • For **Region 2**, the setup is: - The inner integration for \(y\) ranges from \(\ln x\) to 1. - The outer integral, concerning \(x\), stretches from 1 to \(e\).
These limits dictate the boxes "inside" which our integration works. Essentially, they carve out the exact piece of plane we're interested in. Being thorough with bounds ensures proper computation of the double integral. Notably, each limit is directly influenced by the intersecting points of the region's boundaries. Remember, keeping these limits clear not only aids in computation but also affirms conceptual understanding of the integral's span.

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