The area bounded by the curve 2y^2=x and the line 4y=x is rotated around the y-axis. The volume of the resulting structure can be expressed as V=a/bπ, where a and b are coprime positive integers. What is the value of a+b?
5 answers
something's wrong. The region is not closed.
i believe its bounded from y=0 to y=2
Ah. In that case, using discs (washers),
v = ∫[0,2] π(R^2-r^2) dy
where R = 4y and r = 2y^2, so
v = ∫[0,2] π((4y)^2-(2y^2)^2) dy
= 4π∫[0,2] 4y^2 - y^4 dy
= 1024/15 π
v = ∫[0,2] π(R^2-r^2) dy
where R = 4y and r = 2y^2, so
v = ∫[0,2] π((4y)^2-(2y^2)^2) dy
= 4π∫[0,2] 4y^2 - y^4 dy
= 1024/15 π
no, i think that
∫[0,2] π((4y)^2-(2y^2)^2) dy is 256/15π
∫[0,2] π((4y)^2-(2y^2)^2) dy is 256/15π
you may be right. maybe I mixed in an unneeded factor of 4 somewhere.
3 answers