Hello!

The surface area of revolution of the graph of `g ( y )` over the y-axis is computed using definite integral:

`A = int_a^b (2 pi g ( y ) sqrt( 1 + ( g' ( y ) )^2 ) ) dy .`

Here, `y = root ( 3 ) ( 7 x ) , ` so `g(y) = 1 / 7 y^3 ` and `g'(y) = 3 / 7 y^2 . ` Also, `a = root ( 3 ) ( 0 ) = 0, ` `b = root ( 3 ) ( 8 ) = 2 .` This way, the area of revolution is equal to

`int_0^2 ( 2 / 7 pi y^3 sqrt ( 1 + 9/49 y^4 ) ) dy . `

Perform the change of variable `1 + 9 / 49 y^4 = u, ` then `du = 36 / 49 y^3 dy ` and `u ` changes from `1 ` to `1 + 144 / 49 . ` The integral becomes simple:

`int_1^(1 + 144 / 49) ( 49 / 36 * 2 / 7 pi sqrt ( u ) ) du = 7 / 18 pi * 2 / 3 ( u^( 3 / 2 ) )_(u = 1)^( 1 + 144 / 49 ) = `

` = 7 / 27 pi ( ( 1 + 144/49 ) ^ ( 3 / 2 ) - 1 ),`

which is approximately **5.5524**.

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