Question

ok, I'm using the integral formula for surface area to find the surface area of y=x^3, 0 14 years ago

Answer 1

I know the formula but Im doing something wrong I've got \[2\pi \int\limits_{0}^{2} x (u)^1/2 du/36x^3\]

so does the x^3 on the 36x^3 cancel out all my x values

Answer 2

crap hang on let me rewrite that formula

Answer 3

\[2\pi \int\limits_{0}^{2} x(u)^{1/2} du/36x^3\]