Recent content by lila12345

yes the area of that surface! hahah sorry I though it was the right way to say it in English sorry..! and there is no further info in the problem the question is: calculate the area of the surface if you rotate a circle y^2+(x-4)^2=2^2 around the vertical axe (x=0) and again sorry I meant...

HELP I can't find the surface of revolution! By donuts I mean a circle that doesn't touch the axes (tore in french) y^2+(x-4)^2=2^2 is my function ( y^2+x^2=r^2) and the axe of rotation is y so y= sqrt(r^2-x^2) the formula I know : 2* pi (Integral from a to b (F(x)*sqrt( 1+ (f``(x))^2))...