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The area bounded by the curve y = 2x^2-x^3 and line y=0 is rotated around the y-axis. The volume of the resulting structure can...Asked by andy
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?
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Answered by
Steve
something's wrong. The region is not closed.
Answered by
andy
i believe its bounded from y=0 to y=2
Answered by
Steve
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
Answered by
andy
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
Answered by
Steve
you may be right. maybe I mixed in an unneeded factor of 4 somewhere.
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