Answer to Question #131613 in Calculus for Nick

Question

Find the arc length of the curve x=y^4/8+1/4y^2 from y=1 to y=4

Answer 1

$L=∫_1^4 \sqrt{1+((y^4/8+1/(4y^2))')^2}dy =∫_1^4 \sqrt{1+(y^3/2-1/(2y^3))^2}dy=∫_1^4 \sqrt{1+(y^6/4-1/2+1/(4y^6)}dy=\\∫_1^4 \sqrt{1+y^6/4-1/2+1/(4y^6)}dy=∫_1^4 \sqrt{y^6/4+1/2+1/(4y^6)}dy=∫_1^4 \sqrt{(y^3/2+1/(2y^3)^2}dy=\\∫_1^4 (y^3/2+1/(2y^3))dy= (y^4/8-y^{-2})|_1^4=(4^4/8-4^{-2})-(1^4/8-1^{-2})=\\=32-1/16-1/8+1=33-3/16=525/16$