the curve is rotated about the y-axis. find the area of the resulting surface. y=cubert(x), 2 < or equal to y is < or equal to 3

please explain

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You should remember the formula that gives the area of the surface resulted by rotating the curve `f(x) = root(3)x` , around x axis, such that:

`A = 2pi*int_a^b f(x)sqrt(1 + (f'(x))^2)dx`

You need to differentiate the function with respect to x such that:

`f'(x) = (1/3)x^(1/3 - 1) = > f'(x) = (1/3)x^(-2/3) => f'(x) = 1/(3root(3)(x^2))`

You should raise to square `f'(x)` such that:

`(f'(x))^2 = 1/(9x*root(3)x) => 1 + (f'(x))^2 = 1 + 1/(9x*root(3)x) `

`sqrt(1 + (f'(x))^2) = sqrt(1 + 1/(9x*root(3)x)`

`f(x)*sqrt(1 + (f'(x))^2) = (root(3)x)*sqrt(1 + 1/(9x*root(3)x)`

You need to simplify radicals such that:

`(root(3)x)*sqrt((9x*root(3)x + 1)/(9x*root(3)x)) = (sqrt(9x*root(3)x + 1))/(root(3)x)`

You should integrate now such that:

`int (sqrt(9x*root(3)x + 1))/(root(3)x) dx`

You should come up with the substitution such that:

`x^(2/3) = t => dt = 2/(3root(3)x)`

`int (sqrt(9x*root(3)x + 1))/(root(3)x) dx = (3/2)int sqrt(t^2 + 1/9)`

You should come up with the substitution such that:

`t = tan(y)/3 => dt = (1/3)sec^2 y dy`

`(3/2)int sqrt(t^2 + 1/9) = (1/6) int sec^3 y dy`

`(1/6) int sec^3 y dy = (1/12) tan y*sec y + (1/12)log(tan y + sec y) + c`

Substituting back `y = tan^(-1) (3t)` yields:

`(3/2)int sqrt(t^2 + 1/9) = (1/12) tan (tan^(-1) (3t))*sec (tan^(-1) (3t))+ (1/12)log(tan(tan^(-1) (3t)) + sec (tan^(-1) (3t))) + c`

Substituting back `t =x^(2/3)` yields:

`int (sqrt(9x*root(3)x + 1))/(root(3)x) dx = (1/12)(3x^(4/3)sqrt(x^(-4/3)+9) + (x^(2/3)sqrt(x^(-4/3) + 9)sinh^(-1)(3x^(2/3)))/(sqrt(9x^(4/3)+1))) + c`

Hence, you should substitute 3 for b and 2 for a such that:

`A= 2pi*int (sqrt(9x*root(3)x + 1))/(root(3)x) dx = 2pi*1.37752`

**Hence, evaluating the area of surface of revolution, under the given conditions, yields `A = 2.75504 pi` .**

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