What is the area of the surface generated wheny=((x^3)/6)+((1)/2x), from x=1, to x=3?
Last edited by chevy900ss; Aug 2nd 2009 at 02:08 PM.
Follow Math Help Forum on Facebook and Google+
You really need to give more information than that. What object are we finding the surface area of?
ok i rewrote the question.
Ok I think you are after the following equation $\displaystyle SA = 2\pi \int_1^3 f(x) \sqrt{1+[f'(x)]^2} dx$ where $\displaystyle f(x) =\frac{x^3}{6}+\frac{1}{2x}$ can you find $\displaystyle f'(x)$ ?
derivative=((x^2)/2)+((1)/-2x^2)
using that equation i get 214pie/9. Is that right?
Originally Posted by chevy900ss derivative=((x^2)/2)+((1)/-2x^2) Not quite but a good effort. $\displaystyle f'(x) = \frac{x^2}{2}-\frac{2}{x^2} $
Last edited by pickslides; Aug 2nd 2009 at 02:56 PM. Reason: typo
how is it -((2)/(x^2). I thought ((1)/2x) is the same as ((x^-1)/2) where you bring the (-1) down front and subtract (1) from the exponent which gives you -((x^-2)/2). Which is -1/2x^2
After a second look you are correct $\displaystyle f'(x) = \frac{x^2}{2}-\frac{1}{2x^2}$ Sorry about that my friend!
Originally Posted by chevy900ss using that equation i get 214pie/9. Is that right? Ok is this right using that derivative?
View Tag Cloud