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

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

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

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

$pi[72+243/5]$

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

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