How do you find the volume of the solid generated by revolving the region bounded by the graphs y=3(2-x), y=0, x=0, about the y axis?

1 Answer
Dec 30, 2016

Volume = 8pi

Explanation:

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When y(2-x) is rotated about the y-axis it will generate a cone of base radius 2 and height 6, hence the volume is:

\ \ \ \ \ V =1/3pir^2h
:. V = 1/3pi*2^2*6
:. V = 8pi

If you want a calculus solution, then the Volume of revolution about the y-axis is given by:

V =int_a^b pi x^2dy

Now:

y=3(2-x) => y=6-3x
:. x=1/3(6-y)

And so:

V = int_0^6 pi (1/3(6-y))^2 \ dy
\ \ \ = 1/9 pi int_0^6 (6-y)^2 \ dy
\ \ \ = 1/9 pi int_0^6 (36-12y+y^2) \ dy
\ \ \ = 1/9 pi [36y-6y^2+1/3y^3]_0^6
\ \ \ = 1/9 pi {(36*6-6*36+216/3) - (0)}
\ \ \ = 1/9 pi (72)
\ \ \ = 8 pi