Can someone help me answer this thank u The image is an educational diagram discussing the lateral surface area of a cone generated by revolving a line segment around the y-axis. The line segment is defined by \( x = 1 - y \) for \( 0 \leq y \leq 1 \).**Diagram Explanation:**- The illustration shows a cone with its vertex at point \( A(0, 1) \) and a base edge at point \( B(1, 0) \). The line segment \( AB \) is rotated about the y-axis to form the cone.- The equation \( x + y = 1 \) is labeled alongside the line segment, indicating the relationship between \( x \) and \( y \).**Mathematical Expressions:**The problem involves calculating the lateral surface area \( S \) of the cone. Various integral expressions are presented as possible solutions:1. \[ S = 2\pi \int_{c}^{d} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy = 2\pi \int_{0}^{1} (1-y) \sqrt{2y} \, dy \]2. \[ S = 2\pi \int_{c}^{d} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy = 2\pi \int_{0}^{1} (1-y) \sqrt{2} \, dy \]3. \[ S = 2\pi \int_{c}^{d} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy = 2\pi \int_{0}^{1} (1+y) \sqrt{2y} \, dy \]4. \[ S = 2\pi \int_{c}^{d} x \sqrt{1 + \left( \frac{dx}{dy} \right)^2} \, dy = 2\pi \int_{0}^{1} (1-y) \sqrt{y} \, dy \]5. \[ S = 2\pi \int_{c}^{d} x \sqrt{1 + \left( \frac

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The line segment x = 1 - y, 0 ≤ y ≤ 1, is revolved about the y-axis to generate the cone in Figure 6.49. Find its lateral surface area (which excludes the base area). A (0, 1) x+y=1 B (1,0) FIGURE 6.49 Revolving line segment AB about the y-axis generates a cone whose lateral surface area we can now calculate O dx 2 ° S = 2x [^*^ √/ 1 + ( 1 ² ³ dy = 2x / ² (1 S X, 1+ dy 0 (1-y)√2y dy 1 dx 2 S = 2x [*^ √ 1 + = =+) ² dy = 2 = [" (1-3)√2dy S dy 0 dx S = 27 / ²x √/1 + (# ) ² dy = 27 / ² (1+ dy 0 dy=2л (1+y) √2dy O dx S=2x [ ^x√/1+ (#5) ³ dy=25 / ^ (1-y) √ydy dy 0 dx ° 2020 ["^ √/10 (2) 4-2 [ " (2-31 / 54 S = 2n x,/ 1+ dx=2r √ 2dx dy 0

The line segment x = 1 - y, 0 ≤ y ≤ 1, is revolved about the y-axis to generate the cone in Figure 6.49. Find its lateral surface area (which excludes the base area). A (0, 1) x+y=1 B (1,0) FIGURE 6.49 Revolving line segment AB about the y-axis generates a cone whose lateral surface area we can now calculate O dx 2 ° S = 2x [^^ √/ 1 + ( 1 ² ³ dy = 2x / ² (1 S X, 1+ dy 0 (1-y)√2y dy 1 dx 2 S = 2x [^ √ 1 + = =+) ² dy = 2 = [" (1-3)√2dy S dy 0 dx S = 27 / ²x √/1 + (# ) ² dy = 27 / ² (1+ dy 0 dy=2л (1+y) √2dy O dx S=2x [ ^x√/1+ (#5) ³ dy=25 / ^ (1-y) √ydy dy 0 dx ° 2020 ["^ √/10 (2) 4-2 [ " (2-31 / 54 S = 2n x,/ 1+ dx=2r √ 2dx dy 0

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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