`x = 1/3(y^2 + 2)^(3/2) , 0

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,

`x=1/3(y^2+2)^(3/2)`

`dx/dy=1/3(3/2)(y^2+2)^(3/2-1)(2y)`

`dx/dy=y(y^2+2)^(1/2)`

Plug in the above derivative in the arc length formula,

`L=int_0^4sqrt(1+(y(y^2+2)^(1/2))^2)dy`

`L=int_0^4sqrt(1+y^2(y^2+2))dy`

`L=int_0^4sqrt(1+y^4+2y^2)dy`

`L=int_0^4sqrt((y^2+1)^2)dy`

`L=int_0^4(y^2+1)dy`

`L=[y^3/3+y]_0^4`

`L=[4^3/3+4]-[0^3/3+0]`

`L=[64/3+4]`

`L=[(64+12)/3]`

`L=76/3`

Arc length of the function over the given interval is `76/3`