Surface area by revolving a curve problem

[Waggattack]

1. I am suppose to find the surface area of the curve y=sqrt(4-x^2) from -1 to 1 when it is revolving around the x-axis.



2. Homework Equations : S= 2PIf(x)sqrt(1+(dy/dx)^2)dx



3. I found the derivative to be -x(4-x^2)^-1/2 and then squared it so the problem is 2Pi -1$$\int$$1 sqrt(4-x^2)sqrt(1+[x^2(4-x^2)] but I have no clue where to go from here.

 

[Elucidus]

You state that you get

$$2\pi \int_{-1}^{1} \sqrt{4-x^2} \sqrt{1+ \frac{x^2}{4-x^2}} \; dx$$

Try to simplify the integrand. Remember $$\sqrt A \sqrt B = \sqrt{AB}$$.

--Elucidus