How do you find the volume of a solid that is enclosed by #y=x+1#, #y=x^3+1#, x=0 and y=0 revolved about the x axis?

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Answer 1 · 2016-10-21T14:40:29

The volume #=(3pi)/4#

A mall volume (rotation about the x axis #dV=piy^2dx#

We have to find the volume enclosed by the curves
so #dV=piy_1^2dx-piy_2^2dx#

Hence #dV=pi(x^3+1)^2dx-pi(x+1)^2dx#

#dV=pi((x^6+2x^3+1)-(x^2+2x+1))dx#

#dV=pi(x^6+2x^3-x^2-2x)dx#
and the limits of the integral when #y=0#
#0=x+1#=>#x=-1#
and #x=0#

So #V=piint_-1^0(x^6+2x^3-x^2-2x)dx#

#V=pi(x^7/7+(2x^4)/4-x^3/3-(2x^2)/2)_-1^0#

#V=pi(0-(-1/7+1/2+1/3-1))= pi(1/7+1-5/6)#

#V=pi*13/42=(3pi)/4#