Asked by H
                Find the volume of the solid obtained by rotating the region enclosed by the curves y=1x and y=3−x about the x-axis. 
            
            
        Answers
                    Answered by
            Steve
            
    I assume you meant y = 1/x
If so, then the curves intersect at
((3+√5)/2,(3-√5)/2) and ((3-√5)/2,(3+√5)/2)
call these points (a,b) and (b,a)
So, using discs (washers) of thickness dx, we have
v = ∫[a,b] π(R^2-r^2) dx
where R=3-x and r=1/x
= ∫[a,b] π((3-x)^2-1/x^2) dx = 5π√5/3
Or, using shells of thickness dy,
v = ∫[a,b] 2πrh dy
where r=y and h=(3-y)-(1/y)
= ∫[a,b] 2πy((3-y)-(1/y)) dy = 5π√5/3
    
If so, then the curves intersect at
((3+√5)/2,(3-√5)/2) and ((3-√5)/2,(3+√5)/2)
call these points (a,b) and (b,a)
So, using discs (washers) of thickness dx, we have
v = ∫[a,b] π(R^2-r^2) dx
where R=3-x and r=1/x
= ∫[a,b] π((3-x)^2-1/x^2) dx = 5π√5/3
Or, using shells of thickness dy,
v = ∫[a,b] 2πrh dy
where r=y and h=(3-y)-(1/y)
= ∫[a,b] 2πy((3-y)-(1/y)) dy = 5π√5/3
                    Answered by
            Steve
            
    ahem. For shells, it is ∫[b,a]
    
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.