# Find the surface area of revolution about the x-axis of `y=6 sin(2 x)` over the interval `0 le x le pi/2`.

Hello!

The formula for a surface area of revolution is

`A = int _ a ^ b ( 2 pi f( x ) sqrt ( 1 + ( f ' ( x ) ) ^ 2 ) ) dx .`

Here `a = 0 , ` `b = pi / 2 , ` `f ( x ) = 6 sin ( 2 x ), ` so `f ' ( x ) = 12 cos ( 2 x ) ` and the formula becomes

`A = int _ 0 ^ ( pi / 2 ) ( 2 pi * 6 sin ( 2 x ) sqrt ( 1 + 144 cos ^ 2 ( 2 x ) ) ) dx . `

To find this integral, do a variable substitution `u = 12 cos ( 2 x ), ` then `du = - 24 sin ( 2 x ) `, and as `x ` changes from 0 to `pi/2, ` `u ` changes from 12 to -12 (once). Thus, the integral becomes

`int_12^(-12) ( - 1/2 pi sqrt ( 1 + u^2 ) ) du.`

This indefinite integral is more or less a table one,

`int ( - 1/2 pi sqrt ( 1 + u^2 ) ) du = -pi/4 (u sqrt(1+u^2) + sinh^(-1)u)+ C,`

so the definite integral is `A = -pi/4(-12sqrt(145)-12sqrt(145) + sinh^(-1)(-12)-sinh^(-1)(12)) = ` `=pi/2(12sqrt(145)+sinh^(-1)(12)).`

What is `sinh^(-1)(12)?` It is the number `v ` such that `(e^v-e^(-v))/2=12,` or

`(e^v)^2-24e^v-1=0, ` so `e^v=12+sqrt(145)` and `v = ln(1+sqrt(145)).`

This way, the exact answer is `pi/2(12sqrt(145)+ln(1+sqrt(145))).`

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