Surface area on y-axis

[hgaon001]

y=x^3 from 0 to 1 on the y-axis

im guessin u put it in X terms.. so x=y^1/3 then u find the derivative and plug into the formula.. the integral of 2pi (f(y))(sqrt(1+(f'(y)^2)dy ?

 

[DrMike]

Yes, that will work.

 

[BigGlenntheHeavy]

Are you talking of the area of a surface of revolution? If so, then;

$\displaystyle S \ = \ 2\pi\int_{a}^{b}r(x)\sqrt(1+[f ' (x)]^{2})dx$

$\displaystyle In \ your\ \ case \ f(x) \ = \ x^{3}, \ ergo \ \ S \ = \ 2\pi\int_{0}^{1}x\sqrt(1+9x^{4})dx, \ revolved \ around \ y \ axis.$

Note: It is immaterial which axis it is revolved around (horizontal or vertical)

If revolved around y axis r(x) usually is x, if revolved around x axis, r(x) is usually f(x), r(x) being the distance between the graph of f and the axis of revolution.