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
Explanation:
When
\ \ \ \ \ 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
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