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

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The surface area of the solid obtained by rotating the curve `y = sqrt t` about the t-axis for t in[4, 9] is given by the integral `int_4^9 2*pi*y*ds` where `ds = sqrt(1 +(dy/dt)^2) dt`

`y = sqrt t`

`(dy)/(dt) = 1/(2*sqrt t)`

`ds` = `sqrt(1 +(1/(2*sqrt t))^2) dt`

=> `ds = sqrt((4t + 1)/(4t)) dt`

`int_4^9 2*pi*y*ds`

= `int_4^9 2*pi*y*sqrt((4t + 1)/(4t)) dt`

= `int_4^9 2*pi*sqrt t*sqrt((4t + 1)/(4t)) dt`

= `pi*int_4^9 sqrt(4t + 1) dt`

= `(pi/6)*[(4t+1)^(3/2)]_4^9`

= `(pi/6)*(37^(3/2) - 17^(3/2))`

`~~ 81.14`

**The required surface area is 81.14**