# find surface area obtained by rotating `y=sqrt(x), x in[4,9]` about the y-axis.

### 1 Answer | Add Yours

The graph of `y= sqrt x` with `x in [4, 9]` is :

When this is rotated about the y-axis the surface area is `int_4^9 2*pi*x ds` where `ds = sqrt(1 + (dy/dx)^2) dx`

`y = sqrt x` , `dy/dx = 1/(2*sqrt x)`

`ds = sqrt(1 + (1/(2*sqrt x))^2) dx`

`int_4^9 2*pi*x ds`

= `int_4^9 2*pi*x sqrt(1 + (1/(2*sqrt x))^2) dx`

= `int_4^9 2*pi*x sqrt(1 + 1/(4x)) dx`

= `int_4^9 pi*sqrt x sqrt(4x + 1) dx`

= `int_4^9 pi*sqrt (4x^2 + x) dx`

= `(pi/32)*[16*sqrt(4x^4+x^3) + 2*sqrt(4x^2+x)-ln(sqrt(4x+1)+2*sqrt x)]_4^9`

`~~` 6.50286

**The required surface area is approximately 6.50286 square units.**