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3. Consider the region bounded by y = ln x, y = 0, and x = 2. A solid is formed from this base region R whose cross-sections perpendicular to the base and parallel to the x-axis are squares. The following is an INCORRECT solution to the problem of setting up an integral that can be used to find the volume of the solid. Solution: We begin by sketching the base region: 02 OF £2 -42 62 04 46 ** 22 X 24

3. Consider the region bounded by y = ln x, y = 0, and x = 2. A solid is formed from this base region R whose cross-sections perpendicular to the base and parallel to the x-axis are squares. The following is an INCORRECT solution to the problem of setting up an integral that can be used to find the volume of the solid. Solution: We begin by sketching the base region: 02 OF £2 -42 62 04 46 ** 22 X 24

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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3. Consider the region bounded by y = ln x, y = 0, and x = 2. A solid is formed from this base region R whose cross-sections perpendicular to the base and parallel to the x-axis are squares. The following is an INCORRECT solution to the problem of setting up an integral that can be used to find the volume of the solid. Solution: We begin by sketching the base region: 0.2 0.8 0.6 0.4 0.2 0 -0.2 y 02 04 06 Thus, V = 0.8 1 12 14 1.6 22 X 24 Because cross-sections are perpendicular to the y-axis, our variable of integration is y. Note that the region begins when y = 0 and ends when y = ln 2. We rewrite the function y = ln x so that x is a function of y to obtain x = e. Thus, our cross-sectional area is A(y) = (e)² = ey². In 2 110² ev dy. Identify all errors and write a correct solution to the problem.
Transcribed Image Text:3. Consider the region bounded by y = ln x, y = 0, and x = 2. A solid is formed from this base region R whose cross-sections perpendicular to the base and parallel to the x-axis are squares. The following is an INCORRECT solution to the problem of setting up an integral that can be used to find the volume of the solid. Solution: We begin by sketching the base region: 0.2 0.8 0.6 0.4 0.2 0 -0.2 y 02 04 06 Thus, V = 0.8 1 12 14 1.6 22 X 24 Because cross-sections are perpendicular to the y-axis, our variable of integration is y. Note that the region begins when y = 0 and ends when y = ln 2. We rewrite the function y = ln x so that x is a function of y to obtain x = e. Thus, our cross-sectional area is A(y) = (e)² = ey². In 2 110² ev dy. Identify all errors and write a correct solution to the problem.
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