Find the volume of the solid obtained by rotating the region in the first quadrant enclosed by the curves y= tanx and y=2cosx between the bounds 0 and pi/2 around y=-1

would it be

A= pi * ∫ 0 to 0.78 {(2cosx+1)^2 - (tanx+1)^2}

1 answer

yes, if you include dx
The volume consists of washers of thickness dx and area pi(R^2-r^2) as you indicated above