Asked by fatma
Find the exact arc length x=1/3(y^2 +2)^3/2 from y= 0 to y = 1
Answers
Answered by
Steve
x' = y√(y^2+2)
so, the arc length
s = ∫[0,1] √(1+x'^2) dy
= ∫[0,1] √(1+(y^2)(y^2+2)) dy
= ∫[0,1] √(y^4+2y^2+1) dy
= ∫[0,1] √(y^2+1)^2 dy
= ∫[0,1] y^2+1 dy
= 4/3
so, the arc length
s = ∫[0,1] √(1+x'^2) dy
= ∫[0,1] √(1+(y^2)(y^2+2)) dy
= ∫[0,1] √(y^4+2y^2+1) dy
= ∫[0,1] √(y^2+1)^2 dy
= ∫[0,1] y^2+1 dy
= 4/3
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