How do you find the volume of the solid obtained by rotating the region bounded by the curves #Y=2x# and #Y=x^2# rotated around the x-axis?

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Answer 1 · 2015-05-23T10:05:57

Remembering that the area between #f(x)# and #g(x)#, with #f(x)# over #g(x)# from #x_1# to #x_2# is given by:

#int_(x_1)^(x_2)[f(x)-g(x)]dx#,

and since #y=2x# is over the function #y=x^2#, and the intercept point is given by the system:

#y=2x#

#y=x^2#

#2x=x^2rArrx(x-2)=0rArr#

#x_1=0# and #x_2=2#,

then:

#int_0^1(2x-x^2)dx=[x^2-x^3/3]_0^2=#

#=[2^2-2^3/3-(0^2-0^3/3)]=4-8/3=4/3#.