Find the length of the curve x=(y^3)÷3+1÷(4y) from y=1 to y=3?

1 Answer
Ultrilliam
May 6, 2018

# s = 53/6 " units "#

Explanation:

#s = int_(y_1)^(y_2) \ sqrt( 1 + (x')^2) \ dy#

#x' = y^2 - 1/(4 y^2)#

#s = int_(1)^(3) \ sqrt( 1 + y^4 + 1/(16 y^4) - 1/2) \ dy#

# = int_(1)^(3) \ sqrt( y^4 +1/2 + 1/(16 y^4) ) \ dy#

# = int_(1)^(3) \ sqrt( (y^2 + 1/(4 y^2))^2 ) \ dy#

# = ( \ y^3/3 - 1/(4 y) )_1^3 = 53/6 " units "#

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