Asked by tom
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}
would it be
A= pi * ∫ 0 to 0.78 {(2cosx+1)^2 - (tanx+1)^2}
Answers
Answered by
oobleck
yes, if you include dx
The volume consists of washers of thickness dx and area pi(R^2-r^2) as you indicated above
The volume consists of washers of thickness dx and area pi(R^2-r^2) as you indicated above
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