Question
Find the centroid of the region in the first quadrant bounded by the x-axis, the parabola v ^ 2 = 2x , and the line x+y=4
Answers
oobleck
The two curves intersect at (2,2)
the area of the region is
A = ∫[0,2] ((4-y) - y^2/2) dy = 14/3
x̅ = ∫ x dA/A = (3/14)∫[0,2] (y^2/2)((4-y) - y^2/2) dy = (3/14)(26/15) = 13/35
y̅ = ∫ x dA/A = (3/14)∫[0,2] (y)((4-y) - y^2/2) dy = (3/14)(10/13) = 5/7
so the centroid is at (13/35 , 5/7)
the area of the region is
A = ∫[0,2] ((4-y) - y^2/2) dy = 14/3
x̅ = ∫ x dA/A = (3/14)∫[0,2] (y^2/2)((4-y) - y^2/2) dy = (3/14)(26/15) = 13/35
y̅ = ∫ x dA/A = (3/14)∫[0,2] (y)((4-y) - y^2/2) dy = (3/14)(10/13) = 5/7
so the centroid is at (13/35 , 5/7)