How do you find the volume of the solid obtained by rotating the region bounded by the curves #y=x#, #x=0#, and #y=(x^2)-6# rotated around the #y=3#?

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Answer 1 · 2015-08-02T18:55:22

#piint_0^3(3-(x^2-6))^2-(3-x)^2dx=(603pi)/5#

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The grey region is what we will be rotating around the horizontal line #y=3#.

The outer radius is #3-(x^2-6)#

The inner radius is #3-x#

Using the method of washers

#piint_0^3(3-(x^2-6))^2-(3-x^2)^2dx#

#piint_0^3(9-x^2)^2-(3-x)^2dx #

#piint_0^3 81-18x^2+x^4-(9-6x+x^2)dx#

#piint_0^3 81-18x^2+x^4-9+6x-x^2dx#

#piint_0^3 72-19x^2+x^4+6xdx#

Integrating

#pi[72x-19/3x^3+x^5/5+3x^2]#

#pi[72(3)-19/3(3)^3+3^5/5+3(3)^2]#

#pi[216-171+243/5+27]#

#pi[72+243/5]#

#pi[360/5+243/5]=(603pi)/5#