Expert: Socrates - 8/7/2006

Question
I'm trying to set up the equation y=4-x^2, for x=0 to 2 revolved around the y-axis.  The equation i know is S.A. = int (2pi x ds).  Could you help me with it.  I am struggling with the ds part where ds = sqrt (1+(dy/dx)^2).  Thanks,  

Answer
S.A. = 2pi S x( 1 + (dy/dx)^2 )^1/2 dx

dy/dx = -2x

so S.A. = 2pi S x(1+4x^2)^1/2 dx

where the limits for the integral are from x=0 to x=2

An antiderivative for x(1+4x^2)^1/2 will be

(1/12)(1+4x^2)^3/2

so evaluate at x=2 and get (1/12)(17)^3/2
evaluate at x=0 and get 1/12

The integral is then (1/12) ((17)^3/2 - 1)

So the surface area is 2pi times this,

(pi/6) ((17)^3/2 - 1) is the answer.