Answer to Question #154245 in Calculus for mafizur alam

Question

determine the length of x=4(3+y)^2, 1<=y<=4

Answer 1

x = 4(3 + y)²

$\implies$ dx /dy = 4 . 2(3 + y) = 24 + 8y

1+ (dx /dy)² = 1 + (24 + 8y)²

arc length = $\int$₁⁴ $\sqrt{1+ (dx /dy)^2}$ dy = $\int$₁⁴ $\sqrt{1+ (24+8y)^2}$ dy

substituting u = 24 + 8y

$\implies$ du = 8dy , when y =1, u =32 and when y = 4 , u = 56

$\implies$ arc length = $\int$₄⁵⁶ $\sqrt{1+ (u)^2}$ du /8

= [(u$\sqrt{1+ (u)^2}$ + ln(u +$\sqrt{1+ (u)^2}$)/16​​​] ₄⁵⁶

=[(56$\sqrt{1+ (56)^2}$ + ln(56 +$\sqrt{1+ (56)^2}$)/16​​​]-[(32$\sqrt{1+ (32)^2}$ + ln(32 +$\sqrt{1+ (32)^2}$)/16​​​]

=[(56$\sqrt{3137}$ + ln(56 +$\sqrt{3137}$)/16​​​]-[(32$\sqrt{1025}$ + ln(32 +$\sqrt{1025}$)/16​​​]

=132.0349