How do you find the volume bounded by #x^2=4y# and the line x=2y revolved about the x-axis?

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Answer 1 · 2016-06-28T22:19:33

# = (4pi)/15#

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the small element width dx shown will have volume once revolved around the x axis of

#dV = (pi y_2^2 - pi y_1 ^2)dx#

where #x = 2 y_2# and #x^2 = 4 y_1#

#\implies V = pi int_0^2 \ y_2^2 - y_1 ^2 \ dx#

So # V = pi int_0^2 \ x^2/4 - x^4/16 \ dx#

# = pi/16 int_0^2 \ 4x^2 - x^4 \ dx#

# = pi/16 [ (4x^3)/3 - x^5/5]_0^2#

# = pi/16 64/15#

# = (4pi)/15#