The region enclosed by the curves y =x and y = 1 is shown below: Cross section area of the solid perpendicular to y-axis = A(y) A(y) = (2x) = 4x = 4y Now, vohume of the required solid is V = [ A(y)dy = 4 ycy
The region enclosed by the curves y =x and y = 1 is shown below: Cross section area of the solid perpendicular to y-axis = A(y) A(y) = (2x) = 4x = 4y Now, vohume of the required solid is V = [ A(y)dy = 4 ycy
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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In the attached photo, you can see the solution for the volume between the y=x^2 and y = 1. My question is, why is A(y) = (2x)^2 ? I do not understand where the 2x comes from, my belief was that A(y) = x^2 and I just went from there, I am confused as to where the 2x came from.
![The region enclosed by the curves \( y = x^2 \) and \( y = 1 \) is shown below:
Cross-section area of the solid perpendicular to the y-axis is \( A(y) \).
\[ A(y) = (2x)^2 = 4x^2 = 4y \]
Now, the volume of the required solid is:
\[ V = \int_{a}^{b} A(y) \, dy \]
\[ = \int_{0}^{1} 4y \, dy \]
\[ = 2 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc14f58b9-217c-4014-bf1d-96e2d73135c9%2F06c3b0db-86f7-4f0c-9195-8301121914da%2Fbdrlu84_processed.png&w=3840&q=75)
Transcribed Image Text:The region enclosed by the curves \( y = x^2 \) and \( y = 1 \) is shown below:
Cross-section area of the solid perpendicular to the y-axis is \( A(y) \).
\[ A(y) = (2x)^2 = 4x^2 = 4y \]
Now, the volume of the required solid is:
\[ V = \int_{a}^{b} A(y) \, dy \]
\[ = \int_{0}^{1} 4y \, dy \]
\[ = 2 \]
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