x = 1/3sqrt(y)(y-3) , 1

Arc length (L) of the function x=h(y) on the interval [c,d] is given by the formula,
L=int_c^dsqrt(1+(dx/dy)^2)dy  , if x=h(y) and c <=  y <=  d,
Now let's differentiate the function with respect to y,
x=1/3sqrt(y)(y-3)
dx/dy=1/3{sqrt(y)d/dy(y-3)+(y-3)d/dysqrt(y)}
dx/dy=1/3{sqrt(y)(1)+(y-3)1/2(y)^(1/2-1)}
dx/dy=1/3{sqrt(y)+(y-3)/(2sqrt(y))}
dx/dy=1/3{(2y+y-3)/(2sqrt(y))}
dx/dy=1/3{(3y-3)/(2sqrt(y))}
dx/dy=1/3(3)(y-1)/(2sqrt(y))
dx/dy=(y-1)/(2sqrt(y))
Plug in the above derivative in the arc length formula,
L=int_1^4sqrt(1+((y-1)/(2sqrt(y)))^2)dy
L=int_1^4sqrt(1+(y^2-2y+1)/(4y))dy
L=int_1^4sqrt((4y+y^2-2y+1)/(4y))dy
L=int_1^4sqrt((y^2+2y+1)/(4y))dy
L=int_1^4(1/2)sqrt((y+1)^2/y)dy
L=1/2int_1^4(y+1)/sqrt(y)dy
Now let's compute first the indefinite integral by applying integral substitution,
Let u=sqrt(y)
(du)/dy=1/2(y)^(1/2-1)  
(du)/dy=1/(2sqrt(y))
int(y+1)/sqrt(y)dy=int(u^2+1)2du
=2int(u^2+1)du
=2(u^3/3+u) 
substitute back u= sqrt(y)  and add a constant C to the solution,
=2(y^(3/2)/3+sqrt(y))+C
L=[1/2{2(y^(3/2)/3+sqrt(y)}]_1^4
L=[y^(3/2)/3+sqrt(y)]_1^4
L=[4^(3/2)/3+sqrt(4)]-[1^(3/2)/3+sqrt(1)]
L=[8/3+2]-[1/3+1]
L=[(8+6)/3]-[(1+3)/3]
L=[14/3]-[4/3]
L=10/3
Arc length of the function over the given interval is 10/3