Asked by ifi
Finf the volume of the solid generated by revolving the region bounded by the curve. x=2/(y+1), x=0, y=0, y=3, about y-axis.
My ans is pie unit^3. Is it correct? if wrong can i know the right calculation.
My ans is pie unit^3. Is it correct? if wrong can i know the right calculation.
Answers
Answered by
Reiny
since you are rotating about the y-axis
our volume will be of the nature
V = π∫x^2 dy
= π∫(4/(y+1)^2 dy from y = 0 to 3
= π∫(4(y+1)^-2 dy from 0 to 3
= π [ -4(y+1)^-1 ] from 0 to 3
= π [ -4/(y+1) ] from 0 to 3
= π ( -4/4 - (-4/1) ]
= π (-1 + 4 )
= 3π
our volume will be of the nature
V = π∫x^2 dy
= π∫(4/(y+1)^2 dy from y = 0 to 3
= π∫(4(y+1)^-2 dy from 0 to 3
= π [ -4(y+1)^-1 ] from 0 to 3
= π [ -4/(y+1) ] from 0 to 3
= π ( -4/4 - (-4/1) ]
= π (-1 + 4 )
= 3π
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