In advance, i'm really BAD at the whole u-substitution so any advice would be great! also, thanks in advance 
Finding Integrals for Surface Area:
Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis:
y=x^2 , from O<X<2 (less than OR eqaul to), revolve about the X-AXIS
I used the formula for Surface area for revolution about the X-Axis:
S=int(a-b) 2pieF(x) sqrt [1+{f '(x)}^2] dx
so i get:
a=0 b=2 y=x^2 dy/dx=2x
so...
Int(0-2) 2pieX^s sqrt [1+(2x)^2] dx I pull the 2pie through the sign..
2pie Int(0-2) x^2 sqrt [1+(2x0^2] dx = 2pie Int(0-2) x^2 sqrt [1+4x^2]dx
My teacher says try the u-substitution, and then I slap my forehead.
My first guess is:
Let u=1+4x^2, du= 8xdx, but then i cant seem to manipulate that further (i don't see an 8xdx in the integrand in order to make it work)
PLEASE HELP ME!!! thanks!
Finding Integrals for Surface Area:
Set up an integral for the area of the surface generated by revolving the given curve about the indicated axis:
y=x^2 , from O<X<2 (less than OR eqaul to), revolve about the X-AXIS
I used the formula for Surface area for revolution about the X-Axis:
S=int(a-b) 2pieF(x) sqrt [1+{f '(x)}^2] dx
so i get:
a=0 b=2 y=x^2 dy/dx=2x
so...
Int(0-2) 2pieX^s sqrt [1+(2x)^2] dx I pull the 2pie through the sign..
2pie Int(0-2) x^2 sqrt [1+(2x0^2] dx = 2pie Int(0-2) x^2 sqrt [1+4x^2]dx
My teacher says try the u-substitution, and then I slap my forehead.
My first guess is:
Let u=1+4x^2, du= 8xdx, but then i cant seem to manipulate that further (i don't see an 8xdx in the integrand in order to make it work)
PLEASE HELP ME!!! thanks!