`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`