Asked by nameuzer
find the centroid of the area betwen the curves 2y=x^2; y= x^3
Answers
Answered by
mathhelper
Where do they intersect ?
(1/2)x^2 = x^3
x^2 - 2x^3 = 0
x^2(1 - 2x) = 0
x = 0 and x = 1/2
what is the area of the region between?
the area is defined by y = x^2 / 2 - x^3
A = ∫ (x^2/2 - x^3) dx from 0 to 1/2
= (x^3/6 - x^4/4) | from 0 to 1/2
= ( (1/2)^3 /6 - (1/2)^4 / 4 - (0-0)
= 1/48 - 1/64
= 1/192
recall that (x-bar, y-bar) is the centroid, were
x-bar = 1/A ∫ xy dx = 1/A ∫ (x(x^2/2 - x^3) dx from 0 to 1/2
y-bar = 1/A ∫ (1/2)( (x^2/2)^2 - (x^3)^2 ) dx from 0 to 1/2
check you text to make sure I have the right formulas.
lot's of repetitive integration here,
expand first before you integrate and be patient and careful.
(1/2)x^2 = x^3
x^2 - 2x^3 = 0
x^2(1 - 2x) = 0
x = 0 and x = 1/2
what is the area of the region between?
the area is defined by y = x^2 / 2 - x^3
A = ∫ (x^2/2 - x^3) dx from 0 to 1/2
= (x^3/6 - x^4/4) | from 0 to 1/2
= ( (1/2)^3 /6 - (1/2)^4 / 4 - (0-0)
= 1/48 - 1/64
= 1/192
recall that (x-bar, y-bar) is the centroid, were
x-bar = 1/A ∫ xy dx = 1/A ∫ (x(x^2/2 - x^3) dx from 0 to 1/2
y-bar = 1/A ∫ (1/2)( (x^2/2)^2 - (x^3)^2 ) dx from 0 to 1/2
check you text to make sure I have the right formulas.
lot's of repetitive integration here,
expand first before you integrate and be patient and careful.
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