- #1
lila12345
- 2
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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))
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))
- 1) what are the bornes of the integral and how did you find them
- 2)where do you go from there to have the result of 32pi^2