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region in the plane is the triangle with vertices at (0,0), (1,0), and JR (1-x^2) dy dx e R is the region in the xy-plane given by 0 ≤ y ≤ 1,0 ≤ x ≤ y. can evaluate this integral by integrating first with respect to y and are from O to x, and the limits of integration for x are from 0 to 1. [0,1] [0,x] (1-x^2) dy dx rating with respect to y, we get: [0,1] (y - x^2y) | [O,x] dx

region in the plane is the triangle with vertices at (0,0), (1,0), and JR (1-x^2) dy dx e R is the region in the xy-plane given by 0 ≤ y ≤ 1,0 ≤ x ≤ y. can evaluate this integral by integrating first with respect to y and are from O to x, and the limits of integration for x are from 0 to 1. [0,1] [0,x] (1-x^2) dy dx rating with respect to y, we get: [0,1] (y - x^2y) | [O,x] dx

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
ChapterA: Appendix
SectionA.2: Geometric Constructions
Problem 10P: A soda can has a volume of 25 cubic inches. Let x denote its radius and h its height, both in...
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Hi, I have a different answer here, but I don't know which one is correct. . . Can you help me analyze this answer? Thanks

i would use double integrals for this, and would work it out lika that : The region in the plane is the triangle with vertices at (0,0), (1,0), and (1,1). We can set up the double integral as follows: V = S SR (1 – x^2) dy dx where R is the region in the xy-plane given by 0 ≤ y ≤ 1,0 ≤ x ≤ y. We can evaluate this integral by integrating first with respect to y and then with respect to x. The limits of integration for y are from 0 to x, and the limits of integration for x are from 0 to 1. So we have: V = [[0,1] [[0,x] (1-x^2) dy dx Integrating with respect to y, we get: [0,1] (y - x^2 y) |[0,x] dx [0,1] (x - x^3) dx Integrating with respect to x, we get: V = [x^2/2 - x^4/4] | [0,1] V = 1/4 V = V = Therefore, the volume of the solid under the surface z = 1 - x^2 and above the region in the plane given by 0 ≤ y ≤ 1,0 ≤ x ≤ y, is 1/4 cubic units.
Transcribed Image Text:i would use double integrals for this, and would work it out lika that : The region in the plane is the triangle with vertices at (0,0), (1,0), and (1,1). We can set up the double integral as follows: V = S SR (1 – x^2) dy dx where R is the region in the xy-plane given by 0 ≤ y ≤ 1,0 ≤ x ≤ y. We can evaluate this integral by integrating first with respect to y and then with respect to x. The limits of integration for y are from 0 to x, and the limits of integration for x are from 0 to 1. So we have: V = [[0,1] [[0,x] (1-x^2) dy dx Integrating with respect to y, we get: [0,1] (y - x^2 y) |[0,x] dx [0,1] (x - x^3) dx Integrating with respect to x, we get: V = [x^2/2 - x^4/4] | [0,1] V = 1/4 V = V = Therefore, the volume of the solid under the surface z = 1 - x^2 and above the region in the plane given by 0 ≤ y ≤ 1,0 ≤ x ≤ y, is 1/4 cubic units.
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