A 2×3 rectangle has vertices as (0,0), (2,0), (0,3) and (2,3). It rotates 90◦ clockwise about the point (2,0). It then rotates 90◦ clockwise about the point (5,0), then 90◦ clockwise about the point (7,0), and finally, 90◦ clockwise about the point (10,0). (The side originally on the x-axis is now back on the x-axis.) Find the area of the region above the x-axis and below the curve traced out by the point whose initial position is (1,1).
I believe the rectangle is extraneous information and that the only thing that matters is the point(1,1), the centers of rotation, and the piecewise continuous path the initial point takes.
We need to compute the radius of each ark and its start and stop points.
https://www.desmos.com/calculator/izyyiwx6mn
Method 1: calculus
start center end radius equation
(1,1) (2,0) (3,1) √2 (x-2)^2 + y^2 = 2
(3,1) (5,0) (6,2) √5 (x-5)^2 + y^2 = 5
(6,2) (7,0) (9,1) √5 (x-7)^2 + y^2 = 5
(9,1) (10,0) (11,1) √2 (x-10)^2 + y^2 = 2
integral from 1 to 3 of √2-(x-2)^2) dx
= 1 + π/2 ≈ 2.5708
integral from 3 to 6 of √5-(x-5)^2) dx
= 2 + 5π/4 ≈ 5.9270
integral from 6 to 9 of √5-(x-7)^2) dx
= 2 + 5π/4 ≈ 5.9270
integral from 9 to 11 of √2-(x-10)^2) dx
= 1 + π/2 ≈ 2.5708
Total area: 6 + 7π/2 ≈ 16.99557
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Method 2: not calculus
Separate the total area into what you get from connecting the endpoints with straight lines and then add the arc shaped pieces.
The area under the straight lines (the chords of the arcs) is a straightforward 13.
The arcs are each a quarter of a circle minus the triangle between the limbs and the chord.
πr^2/4 - r^2/2 = (π-2)r^2/4
r^2 is 2,5,5,2; so (π-2)14/4
Total area: 13 + 7π/2 - 7 = 6 + 7π/2 ≈ 16.99557
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Posted by Larry
on 2025-02-12 16:15:25 |
2 methods
